MCQ Questions : The Remainder Theorem and The Factor Theorem

မှန်သော အဖြေကို ရွေးပေးရန် ဖြစ်ပါသည်။

1. The degree of a polynomial is the _____ A. largest coefficient of $x$ B. smallest coefficient of $x$ C. highest power of $x$ D. lowest power of $x$ Explanation ကိန်းရှင် $x$ ပါဝင်သော polynomial တစ်ခုတွင် $x$ ၏ အကြီးဆုံးထပ်ညွှန်းကို ၎င်း polynomial ၏ degree (သို့) order ဟု ခေါ်သည်။ When $9 x^{2}-6 x+2$ is divided by $x-3$, the remainder will be A. $60$ B. $\dfrac{15}{2}$ C. $\dfrac{19}{5}$ D. $65$ Explanation Let $f(x)=9 x^{2}-6 x+2$. When $f(x)$ is divided by $x-3$, The remainder = $f(3)=9 (3)^{2}-6 (3)+2=65$ If a polynomial $f(x)$ is divided by a linear divisor $(x-a)$, then the remainder is A. $f(-a)$ B. $f(a)$ C. $a$ D. $f(0)$ Explanation Remainder Theorem ကို မေးထားခြင်း ဖြစ်ပါသည်။ For what value of $k, x+4$ is a factor of the polynomial $x^{2}-x-(2 k-2) ?$ A. $7$ B. $-9$ C. $11$ D. $-5$ Explanation $x+4$ is a factor of the polynomial $x^{2}-x-(2 k-2)$. By factor theorem, $(-4)^{2}-(-4)-(2 k-2)=0$ $16+4-2(k-1)=0$ $2(k-1)=20$ $k=11$ If the two roots of the equation $x^{3}-x^{2}-5 x+5=0$ are $\sqrt{5}$ and $-\sqrt{5}$, then the third root is A. $1$ B. $-1$ C. $2$ D. $-2$ Explanation The two roots of the equation $x^{3}-x^{2}-5 x+5=0$ are $\sqrt{5}$ and $-\sqrt{5}$. Let the third root be $k$. $\therefore \quad x^{3}-x^{2}-5 x+5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\left(x-k\right)$. $\therefore\quad \sqrt{5}\times \left(-\sqrt{5} \right)\times (-k)=5$ $\quad\quad 5k=5$ $\quad\quad k=1$ What should be subtracted to the polynomial $x^{2}-16 x+30$, so that $x-15$ is the factor of the resulting polynomial? A. $30$ B. $14$ C. $15$ D. $16$ Explanation Let $f(x) = x^{2}-16 x+30$. When $f(x)$ is divided by $x-15$, The remainder = $f(15) = (15)^{2}-16 (15)+30 = 15(15 - 16 + 2) = 15$ Hence, $15$ should be subtracted. The quadratic polynomial $a x^{2}+b x+c=0$ where $c=\dfrac{b^{2}}{4 a}$ has _____real roots. A. $1$ B. $2$ C. $0$ D. None of these. Explanation $a x^{2}+b x+c=0$ where $c=\dfrac{b^{2}}{4 a}$ $\therefore\quad 4ac = b^2$ $\therefore\quad b^2-4ac = 0$ $\therefore\quad a x^{2}+b x+c=0$ has only one root. The remainder when $x^{4}-3 x^{2}+4 x+1$ is divided by $x^{2}-x-1$ is A. $x$ B. $2x$ C. $3x$ D. $4x$ Explanation If the sum and the product of the roots of $x^{2}+b x+c=0$ are $3$ and $2$ , then A. $b=-3, c=2$ B. $b=3, c=2$ C. $b=-3, c=-2$ D. $b=3, c=-2$ Explanation The sum and the product of the roots of $x^{2}+b x+c=0$ are $3$ and $2$. Let the roots be $p$ and $q$. $\therefore\quad p+q = 3, \ pq=2$ $\therefore\quad x^{2}+b x+c=(x-p)(x-q)$ $\therefore\quad x^{2}+b x+c=x^{2}-(p+q) x+pq$ $\therefore\quad x^{2}+b x+c=x^{2}-3x+2$ $\therefore\quad b = -3, \ c = 2$ Given that $f(x)=x^{4}+x^{2}+3 x-1$ and $g(x)=x^{2}-1$. If $f(x)=q(x) g(x)+r(x)$, which of the following is true? $\mathbf{I.}\ q(x)=x^2+2 \quad$ $\mathbf{II.}\ r(x)=3 x+1 \quad$ $\mathbf{III.}\ g(x)$ is a factor of $f(x)$ A. I only B. II only C. I and II only D. I and III only Explanation $f(x)=x^{4}+x^{2}+3 x-1$ and $g(x)=x^{2}-1$ $\quad\quad f(x)=q(x) g(x)+r(x)$ (given) $\therefore\quad f(x)=(x^2+2)(x^2-1)+3x+1$ $\therefore\quad q(x)=x^2+2, \ r(x) = 3x+1$ If $x-3$ is a factor of $x^{3}-(2+k) x^{2}+7 k$, then $k=$ A. $4$ B. $4.5$ C. $5$ D. $5.5$ Explanation $x-3$ is a factor of $x^{3}-(2+k) x^{2}+7 k$. $\therefore\quad 3^{3}-(2+k) (3)^{2}+7 k=0$ $\therefore\quad 27-9\left(2+k\right)+7k=0$ $\therefore\quad k=\dfrac{9}{2} = 4.5$ If $f(x)=6 x^{3}+13 x^{2}+p x+q$ is exactly divisible $2 x^{2}+7 x-4$, then A. $3 x-4$ is a factor of $f(x)$ B. $3 x+4$ is a factor of $f(x)$ C. $4 x-3$ is a factor of $f(x)$ D. $4 x+3$ is a factor of $f(x)$ Explanation $f(x)=6 x^{3}+13 x^{2}+p x+q$ is exactly divisible $2 x^{2}+7 x-4$ $2 x^{2}+7 x-4=(2x-1)(x+4)$ $\therefore\quad 6 \left(\dfrac{1}{2}\right)^{3}+13 \left(\dfrac{1}{2}\right)^{2}+p \left(\dfrac{1}{2}\right)+q=0 \ldots(1)$ $\quad\quad 6 \left(-4\right)^{3}+13 \left(-4\right)^{2}+p \left(-4\right)+q=0\ldots(2)$ $\therefore\quad p=-40,\ q=16$ $\therefore\quad f(x)=6 x^{3}+13 x^{2}-40x+16$ $\therefore\quad f \left(\dfrac{4}{3}\right)=6 \left(\dfrac{4}{3}\right)^{3}+13 \left(\dfrac{4}{3}\right)^{2}-40\left(\dfrac{4}{3}\right)+16$ $\hspace{2.7cm} =\dfrac{128}{9}+\dfrac{208}{9}-\dfrac{160}{3}+16$ $\hspace{2.7cm} =0$ $\therefore\quad 3 x-4$ is a factor of $f(x)$. If $x^{2}+2 x-3$ is a factor of $f(x)=x^{4}+2 x^{3}-7 x^{2}+a x+b$, then A. $a=8$ and $b=12$ B. $a=-8$ and $b=12$ C. $a=8$ and $b=10$ D. $a=-8$ and $b=10$ Explanation $x^{2}+2 x-3$ is a factor of $f(x)=x^{4}+2 x^{3}-7 x^{2}+a x+b$ Let $f(x) = g(x)(x^{2}+2 x-3)$ $\therefore x^{4}+2 x^{3}-7 x^{2}+a x+b = g(x)(x+3)(x-1)$ When $x=-3,\ (-3)^{4}+2 (-3)^{3}-7 (-3)^{2}+a (-3)+b = 0$ When $x=1,\ (1)^{4}+2 (1)^{3}-7 (1)^{2}+a (1)+b = 0$ $\therefore\quad 3a - b =-36 \ \ldots(1)$ $\quad\quad a + b = 4 \ \ \quad\ldots(2)$ $\therefore\quad a= -8, b=12$ If $x^{2}+a x+b$ is divided by $x+c$, then remainder is A. $c^{2}-a c-b$ B. $-c^{2}-a c+b$ C. $c^{2}+a c+b$ D. $c^{2}-a c+b$ Explanation $x^{2}+a x+b$ is divided by $x+c$, then the remainder $= (-c)^{2}+a (-c) +b = c^2-ac+b$ If $x^{3}+9 x+5$ is divided by $x$, then the remainder is A. $9$ B. $5$ C. $-9$ D. $0$ Explanation $x^{3}+9 x+5$ is divided by $x$, then the remainder $= (0)^{3}+9 (0)+5 = 5$ If $(x-1)$ is a factor of $a x-a$, then the value of $a$ is A. $1$ B. $2$ C. $3$ D. any real number Explanation Let $f(x) = ax - a$ $f(1) = a - a = 0$ $\therefore\quad x-1$ is a factor of $f(x)$ for all value of $a\in \mathbb{R}$. What is the remainder when $f(x)=x^{3}+3 x^{2}+k x+k$ is divided by $p x$ ? A. $f(p)$ B. $f\left(\dfrac{1}{p}\right)$ C. $k$ D. $f(k)$ Explanation When $f(x)=x^{3}+3 x^{2}+k x+k$ is divided by $p x$, The remainder = $f(0) = k$ If $x^{2}-3 x+2$ is a factor of the polynomial $f(x)$, then $f(2)=$ A. $0$ B. $1$ C. $2$ D. None of these Explanation $x^{2}-3 x+2$ is a factor of the polynomial $f(x)$. $x^{2}-3 x+2 = (x-1)(x-2)$ $\therefore\quad (x-1)$ and (x-2) are factors $f(x)$. $\therefore\quad f(2)=0$ Given that $f(x)=q(x)\left(x^{2}-4\right)+3$. What is the remainder when $f(x)$ is divided by $x-2$ ? A. $1$ B. $3$ C. $2$ D. $0$ Explanation $f(x)=q(x)\left(x^{2}-4\right)+3$. When $f(x)$ is divided by $x-2$, The remainder $f(2)=q(2)\left(2^{2}-4\right)+3 =3$ If $n$ is a positive integer, what is the remainder when $5 x^{2 n+1}+10 x^{2 n}-3 x^{2 n-1}+5$ is divided by $x+1$. A. $13$ B. $-1$ C. $5$ D. $10$ Explanation Let $f(x)=5 x^{2 n+1}+10 x^{2 n}-3 x^{2 n-1}+5$ where $n$ is a positive integer. When $f(x)$ is divided $x + 1$, The remainder $= f(-1)$. $\hspace{2.5cm} =5 (-1)^{2 n+1}+10 (-1)^{2 n}-3 (-1)^{2 n-1}+5$ $\hspace{2.5cm} =5 (-1)^{\text{odd integer}}+10 (-1)^{\text{even integer}}-3 (-1)^{\text{odd integer}}+5$ where $n\in \mathbb{J^+}$ $\hspace{2.5cm} =5 (-1)+10 (1)-3 (-1) + 5$ $\hspace{2.5cm} =13$ Your Score:

Practice Problems : The Remainder and Factor Theorem

1.        What number should be added to $ \displaystyle 2x^3 - 3x^2 - 8x$ so that the resulting polynomial leaves the remainder $ \displaystyle 10$ when divided by $ \displaystyle 2x + 1$?

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Let the number to be added be $ \displaystyle k$ and the resulting polynomial be $ \displaystyle f(x)$. $ \displaystyle \therefore \ f(x)=2{{x}^{3}}-3{{x}^{2}}-8x+k$ When $ \displaystyle f(x)$ is divided by $ \displaystyle 2x+1$, the remainder is 10. $ \displaystyle \begin{array}{l}\therefore \ f\left( {-\displaystyle \frac{1}{2}} \right)=10\\\\\ \ 2{{\left( {-\displaystyle \frac{1}{2}} \right)}^{3}}-3{{\left( {-\displaystyle \frac{1}{2}} \right)}^{2}}-8\left( {-\displaystyle \frac{1}{2}} \right)+k=10\\\\\therefore \ -\displaystyle \frac{1}{4}-\displaystyle \frac{3}{4}+4+k=10\\\\\therefore \ k=7\end{array}$ Hence, the number to be added is 10.

2.        What number should be subtracted from $ \displaystyle 6x^3 + 7x^2 - 9x+12$ so that $ \displaystyle 3x - 1$ is the factor of the resulting polynomial?

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Let the number to be subtracted be $ \displaystyle k$ and the resulting polynomial be $ \displaystyle f(x)$. $ \displaystyle \therefore \ f(x)=6x^3 + 7x^2 - 9x+12-k$ Since $ \displaystyle 3x - 1$ is the factor of $ \displaystyle f(x)$, $ \displaystyle \begin{array}{l}\ \ f\left( {\displaystyle \frac{1}{3}} \right)=0\\\\\ \ 6{{\left( {\displaystyle \frac{1}{3}} \right)}^{3}}+7{{\left( {\displaystyle \frac{1}{3}} \right)}^{2}}-9\left( {\displaystyle \frac{1}{3}} \right)+12-k=0\\\\\therefore \ \displaystyle \frac{2}{9}+\displaystyle \frac{7}{9}-3+12-k=0\\\\\therefore \ k=10\end{array}$ Hence, the number to be subtracted is 10.

3.        When divided by $ \displaystyle x - 3$ the polynomials $ \displaystyle x^3 - px^2 + x + 6$ and $ \displaystyle 2x^3 - x^2 - (p + 3) x - 6$ leave the same remainder. Find the value of $ \displaystyle p$.

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Let $ \displaystyle f(x)={{x}^{3}}-p{{x}^{2}}+x+6$ and $ \displaystyle g(x)=2x^3 - x^2 - (p + 3) x - 6$ $ \displaystyle f(x)$ and $ \displaystyle g(x)$ leave the same remainder when divided by $ \displaystyle x - 3$, $ \displaystyle \begin{array}{l}\therefore \ f(3)=g(3)\\\\\ \ \ {{(3)}^{3}}-p{{(3)}^{2}}+(3)+6=2{{(3)}^{3}}-{{(3)}^{2}}-(p+3)(3)-6\\\\\ \ \ 27-9p+9=54-9-3p-9-6\\\ \ \\\therefore \ 36-9p=30-3p\ \\\\\therefore \ p=1\ \end{array}$

4.       Using remainder theorem, find the value of $ \displaystyle a$ if the division of $ \displaystyle x^3 + 5x^2 - ax + 6$ by $ \displaystyle x -1$ leaves the remainder $ \displaystyle 2a$.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+5{{x}^{2}}-ax+6\\\\\ \ \ \text{When}\ f(x)\ \text{is divided by }x-1,\ \\\\\ \ \ \text{the remainder}\ =2a\\\\\therefore \ f(1)=2a\\\\\ \ \ {{1}^{3}}+5{{(1)}^{2}}-a(1)+6=2a\\\\\ \ \ 1+5-a+6=2a\\\\\therefore \ 3a=12\\\\\ \ \ a=4\ \ \ \end{array}$

5.        Find the value of the constants $ \displaystyle a$ and $ \displaystyle b$, if $ \displaystyle x - 2$ and $ \displaystyle x + 3$ are both factors of the expression $ \displaystyle x^3 + ax^2 + bx - 12$.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+a{{x}^{2}}+bx-12\\\\\ \ \ (x-2)\ \text{and}\ (x+3)\ \text{are factors of }f(x).\ \\\\\therefore \ f(2)=0\\\\\ \ \ {{(2)}^{3}}+a{{(2)}^{2}}+b(2)-12=0\\\\\therefore \ \ 8+4a+2b-12=0\\\\\therefore \ 2a+b=2---(1)\\\\\ \ \text{Again}\ f(-3)=0\\\\\ \ \ {{(-3)}^{3}}+a{{(-3)}^{2}}+b(-3)-12=0\\\\\therefore \ \ -27+9a-3b-12=0\\\\\therefore \ \ 3a-b=13---(2)\\\\\ \ \ (1)+(2)\Rightarrow 5a=15\\\\\therefore \ \ a=3\\\\\therefore \ 2\left( 3 \right)+b=2\\\\\therefore \ b=-4\end{array}$

6.        If $ \displaystyle x + 2$ and $ \displaystyle x - 3$ are factors of $ \displaystyle x^3 + ax + b$, find the values of $ \displaystyle a$ and $ \displaystyle b$. With these values of $ \displaystyle a$ and $ \displaystyle b$, factorise the given expression.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+ax+b\\\\\ \ \ (x+2)\ \text{and}\ (x-3)\ \text{are factors of }f(x).\ \\\\\therefore \ f(-2)=0\\\\\ \ \ {{(-2)}^{3}}+a(-2)+b=0\\\\\therefore \ \ -8-2a+b=0\\\\\therefore \ -2a+b=8---(1)\\\\\ \ \text{Again}\ f(3)=0\\\\\ \ \ {{(3)}^{3}}+a(3)+b=0\\\\\therefore \ \ 27+3a+b=0\\\\\therefore \ \ 3a+b=-27---(2)\\\\\ \ \ (2)-(1)\Rightarrow 5a=-35\\\\\therefore \ \ a=-7\\\\\therefore \ \ -2\left( {-7} \right)+b=8\\\\\therefore \ \ b=-6\\\\\therefore \ \ f(x)={{x}^{3}}-7x-6\\\\\ \ \ \ \text{Let}\ {{x}^{3}}-7x-6=(x+2)(x-3)(x+k)\\\\\therefore \ \ {{x}^{3}}-7x-6={{x}^{3}}+\left( {k-1} \right){{x}^{2}}-\left( {k+6} \right)x-6k\\\\\therefore \ k-1=0\Rightarrow k=1\\\\\therefore \ \ f(x)=(x+2)(x-3)(x+1)\\\ \ \end{array}$

7.        Given that $ \displaystyle x - 2$ is a factor of the expression $ \displaystyle x^3 + ax^2 + bx + 6$. When this expression is divided by $ \displaystyle x - 3$, it leaves the remainder 3. Find the values of $ \displaystyle a$ and $ \displaystyle b$.

8.        If $ \displaystyle x - 2$ is a factor of the expression $ \displaystyle 2x^3 + ax^2 + bx - 14$ and when the expression is divided by $ \displaystyle x - 3$, it leaves a remainder $ \displaystyle 52$, find the values of $ \displaystyle a$ and $ \displaystyle b$.

9.        If $ \displaystyle ax^3 + 3x^2 + bx - 3$ has a factor$ \displaystyle 2x + 3$ and leaves remainder -3 when divided by (x + 2), find the values of $ \displaystyle a$ and $ \displaystyle b$. With these values of $ \displaystyle a$ and $ \displaystyle b$, factorise the given expression.

10.      Given $ \displaystyle f (x) = ax^2 + bx + 2$ and $ \displaystyle g (x) = bx^2 + ax + 1$. If $ \displaystyle x - 2$ is a factor of $ \displaystyle f (x)$ but leaves the remainder $ \displaystyle -15$ when it divides $ \displaystyle g (x)$, find the values of $ \displaystyle a$ and $ \displaystyle b$. With these values of $ \displaystyle a$ and $ \displaystyle b$, factorise the expression $ \displaystyle f (x) + g (x) + 4x^2 + 7x$.

11.     When $ \displaystyle x^3 - 2x^2 + px - q$ is divided by $ \displaystyle x^2 - 2x - 3$, the remainder is $ \displaystyle x - 6$, What are the values of $ \displaystyle p$ and $ \displaystyle q$ respectively ?

12.      If $ \displaystyle x + k$ is a common factor of $ \displaystyle x^2 + px + q$ and $ \displaystyle x^2 + lx + m$, Find the value of $ \displaystyle k$ in terms of $ \displaystyle p, q, l$ and $ \displaystyle m$.

13.      When a polynomial $ \displaystyle f(x)$ is divided by $ \displaystyle x - 3$ and $ \displaystyle x + 6$, the respective remainders are $ \displaystyle 7$ and $ \displaystyle 22$. What is the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle (x - 3) (x + 6)$ ?

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)=Q(x)(x-3)(x+6)+ax+b\\\\\ \ \text{where is }Q(x)\ \text{a quotient and }ax+b\ \text{is }\\\ \ \text{the remainder when }f(x)\ \text{is divided by}\ (x-3)(x+6).\\\\\ \ \ \text{By the problem,}\\\\\ \ \ f(3)=7\\\\\ \ \ 3a+b=7\ ---(1)\\\\\ \ f(-6)=22\\\\\ \ \ -6a+b=22\ ---(2)\\\\\ \ \ (1)-(2)\Rightarrow 9a=-\displaystyle \frac{5}{3}\\\\\therefore \ 3\left( {-\displaystyle \frac{5}{3}} \right)+b=7\\\\\ \ \ b=12\\\\\therefore \ \text{The remainder when }f(x)\ \text{is divided by}\ \\\ \ \ (x-3)(x+6)\ \text{is}\ -\displaystyle \frac{5}{3}x+12.\end{array}$ $ \displaystyle (x-3)(x+6)$ သည္ polynomial of second degree ျဖစ္ပါသည္။ $ \displaystyle f(x)$ ကို $ \displaystyle (x-3)(x+6)$ ႏွင့္ စားေသာ အႂကြင္းသည္ စားကိန္းေအာက္ တစ္ထပ္ေလ်ာ့ပါမည္။။ ထို႔ေၾကာင့္ အႂကြင္းသည္ $ \displaystyle ax + b$ ပံုစံျဖစ္ပါမည္။

14.      Given that $ \displaystyle f (x) = x^3 + ax^2 + bx + c$. If $ \displaystyle f (1) = f(2) = 0$ and $ \displaystyle f(4) = f(0)$, find $ \displaystyle a, b$ and $ \displaystyle c$.

15.      If $ \displaystyle x – 1$ is a factor of $ \displaystyle Ax^3 + Bx^2 - 36x + 22$ and $ \displaystyle 2^B = 64^A$, find $ \displaystyle A$ and $ \displaystyle B$.

16.      When k is subtracted from $ \displaystyle 27x^3 - 9x^2 - 6x - 5$, it is exactly divisible by $ \displaystyle 3x - 1$, find $ \displaystyle k$.

17.      If $ \displaystyle x^3 + px + q$ and $ \displaystyle x^3 + qx + p$ have a common factor, show that $ \displaystyle p + q +1 = 0$.

18.      Given that $ \displaystyle f(x)=4x^3-4kx^2-x+k$, $ \displaystyle g(x)=3x^2+(1-3k)x-k$ and $ \displaystyle h(x)=f(x)+g(x)$. If $ \displaystyle x-2$ is a factor of $ \displaystyle h(x)$, find the value of If $ \displaystyle k$ and hence solve the equation $ \displaystyle h(x)=0$.

19.      What must be added to $ \displaystyle 6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 1$, so that the polynomial so obtained is exactly divisible by $ \displaystyle 3x^2 - 2x + 4$?

20.      Determine the value of $ \displaystyle p$ for which the polynomial $ \displaystyle 5x^3 - x^2 + 4x+ p$ is divisible by $ \displaystyle 1 - 5x$.

Practice :The Remainder Theorem and The Factor Theorem

1.       It is given that $ \displaystyle f(x)=x^3+ax^2+bx-48$. When $ \displaystyle f(x)$ is divided by $ \displaystyle x - 3$ the remainder is $ \displaystyle 6$. Given that $ \displaystyle f'(1)=0$ , find the value of $ \displaystyle a$ and of $ \displaystyle b$.

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       $ \displaystyle f(x)=x^3+ax^2+bx-48$

       When $ \displaystyle f(x)$ is divided by $ \displaystyle x-3$, the remainder is $ \displaystyle 6$.

$ \displaystyle \begin{array}{l}\therefore \ \ f(3)=6\\\\\ \ \ \ {{(3)}^{3}}+a{{(3)}^{2}}+b(3)-48=6\\\\\ \ \ \ 3a+b=9\,\ \ -----(1)\\\\\ \ \ \ {f}'(x)=3{{x}^{2}}+2ax+b\\\\\ \ \ \ {f}'(1)=0\ \ \ \ [\text{given}]\\\\\ \ \ \ 3{{(1)}^{2}}+2a(1)+b=0\\\\\ \ \ \ 2a+b=-3\,\ \ ----(2)\end{array}$

      By equation (1) - equation (2),

$ \displaystyle \begin{array}{l}\ \ \ \ \ a=12\\\\\therefore \ \ \  24+b=-3\, \Rightarrow b=-27\end{array}$

2.       If $ \displaystyle f(x) = ax^2 + bx + c$ leaves remainders $ \displaystyle 1 , 25 , 1$ on division by $ \displaystyle x - 1 , x + 1 , x - 2$ respectively, show that $ \displaystyle f(x)$ is a perfect square.

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$ \displaystyle \begin{array}{l}\ \ \ f(x)=a{{x}^{2}}+bx+c\\\\\ \ \ f(1)=1\\\\\ \ \ a{{(1)}^{2}}+b(1)+c=1\\\\\ \ \ a+b+c=1\ \ \ \ ----(1)\\\\\ \ \ f(-1)=25\\\\\ \ \ a{{(-1)}^{2}}+b(-1)+c=1\\\\\ \ \ a-b+c=25\ ----(2)\\\\\ \ \ f(2)=1\\\\\ \ \ a{{(2)}^{2}}+b(2)+c=1\\\\\ \ \ 4a+2b+c=1\ \ ---(3)\\\\\ \ \ \text{By equation (1)}-\text{equation (2),}\\\\\ \ \ 2b=-24\Rightarrow b=-12\\\\\ \ \ \text{Substituting }b=-12\ \text{in equations (1)}\operatorname{and}\text{(3),}\\\\\ \ \ a-12+c=1\ \Rightarrow a+c=13\ \ \ \ \ --(4)\\\\\ \ \ 4a-24+c=1\ \Rightarrow 4a+c=25--(5)\\\\\ \ \ \text{By equation (5)}-\text{equation (4),}\\\\\ \ \ 3a=12\Rightarrow a=4\\\\\ \ \ \text{Substituting }a=4,b=-12\ \text{in equations (1), }\\\\\ \ \ 4-12+c=1\ \Rightarrow c=9\\\\\therefore f(x)=4{{x}^{2}}-12x+9={{(2x-3)}^{2}}\\\\\therefore f(x)\ \text{is a perfect square}\text{.}\end{array}$

3.       It is given that $ \displaystyle x-2$ is a factor of $ \displaystyle f(x)=x^3+kx^2-8x-8$ where $ \displaystyle k$ is an integer.

(i) Find the value of the integer $ \displaystyle k$.

(ii) Using your of $ \displaystyle k$, find the non-integer roots of the equation $ \displaystyle f(x)=0$ in the form $ \displaystyle a\pm \sqrt{b}$, where $ \displaystyle a$ and $ \displaystyle b$ are integers.

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$ \displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{3}}+k{{x}^{2}}-8x-8\\\\\ \ \ \ \text{Since}\ x-2\ \text{is a factor of }f(x),\\\\\ \ \ \ f(2)=0\\\\\therefore \ \ {{(2)}^{3}}+k{{(2)}^{2}}-8(2)-8=0\\\\\therefore \ \ 8+4k-16-8=0\\\\\therefore \ \ k=4\\\\\therefore \ \ f(x)={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ \text{Let}\ f(x)=(x-2)({{x}^{2}}+ax+b)\\\\\therefore \ \ (x-2)({{x}^{2}}+ax+b)={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ {{x}^{3}}+(a-2){{x}^{2}}+(b-2a)x-2b={{x}^{3}}+4{{x}^{2}}-8x-8\\\\\ \ \ \ a-2=4\Rightarrow a=6\\\\\ \ \ \ 2b=8\Rightarrow b=4\\\\\therefore \ \ f(x)=(x-2)({{x}^{2}}+6x+4)\\\\\ \ \ \ \text{When }f(x)=0,\\\\\ \ \ \ (x-2)({{x}^{2}}+6x+4)=0\\\\\ \ \ \ (x-2)({{x}^{2}}+6x+9-5)=0\\\\\ \ \ \ (x-2)\left[ {{{{(x+3)}}^{2}}-5} \right]=0\\\\\ \ \ \ x=2\ \text{or }{{(x+3)}^{2}}=5\\\\\therefore \ \ x=2\ \text{or }x=-3\pm \sqrt{5}\\\\\therefore \ \ \text{The non-integer roots of }f(x)=0\ \text{are }-3\pm \sqrt{5}.\end{array}$

4.       A function f is such that $ \displaystyle f(x)=4x^3+4x^2+ax+b$. It is given that $ \displaystyle 2x−1$ is a factor of both $ \displaystyle f(x)$ and $ \displaystyle f'(x)$.

(i) Find the value of $ \displaystyle a$ and $ \displaystyle b$.

(ii) Hence find the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x+3$.

(iii) Express $ \displaystyle f(x)$ in the form $ \displaystyle f(x)=(2x−1)(px^2+qx+r)$, where $ \displaystyle p, q$ and $ \displaystyle r$ are integers to be found.

(iv) find the values of $ \displaystyle x$ for which $ \displaystyle f(x) = 0$.

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$ \displaystyle \begin{array}{l}\ \ \ \ f(x)=4{{x}^{3}}+4{{x}^{2}}+ax+b\\\\\therefore \ \ {f}'(x)=12{{x}^{2}}+8x+a\\\\\ \ \ \ 2x-1\ \text{is a factor of both}\ f(x)\ \text{and}\ {f}'(x).\end{array}$

$ \displaystyle \therefore \ \ f\left( {\frac{1}{2}} \right)=0$

$ \displaystyle \therefore \ \ 4{{\left( {\frac{1}{2}} \right)}^{3}}+4{{\left( {\frac{1}{2}} \right)}^{2}}+a\left( {\frac{1}{2}} \right)+b=0$

$ \displaystyle \ \ \ \ \frac{1}{2}+1+\frac{a}{2}+b=0$

$ \displaystyle \therefore \ \ \ a+2b=-3$

$ \displaystyle \ \ \ \ \ \text{And}\ {f}'\left( {\frac{1}{2}} \right)=0$

$ \displaystyle \therefore \ \ \ 12{{\left( {\frac{1}{2}} \right)}^{2}}+8\left( {\frac{1}{2}} \right)+a=0$

$ \displaystyle \begin{array}{l}\therefore \ \ \ a=-7\\\\\therefore \ \ \ -7+2b=-3\ \Rightarrow b=2\\\\\therefore \ \ \ f(x)=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\ \ \ \ \ \text{When}\ f(x)\ \text{is divided by}x+3,\\\\\ \ \ \ \ \text{the remainder is}\ f(-3).\\\\\therefore \ \ \ f(-3)=4{{(-3)}^{3}}+4{{(-3)}^{2}}-7(-3)+2=-49\\\\\ \ \ \ \ \text{When}\ f(x)=(2x-1)(p{{x}^{2}}+qx+r),\\\\\ \ \ \ \ \ (2x-1)(p{{x}^{2}}+qx+r)=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\ \ \ \ \ \ 2p{{x}^{3}}+(2q-p){{x}^{2}}+(2r-q)x-r=4{{x}^{3}}+4{{x}^{2}}-7x+2\\\\\therefore \ \ \ \ r=-2\\\\\ \ \ \ \ \ 2r-q=-7\Rightarrow -4-q=-7\Rightarrow q=3\\\\\ \ \ \ \ \ 2q-p=4\Rightarrow 6-p=4\Rightarrow p=2\\\\\therefore \ \ \ \ f(x)=(2x-1)(2{{x}^{2}}+3x-2)\\\\\ \ \ \ \ \text{When}\ f(x)=0,(2x-1)(2{{x}^{2}}+3x-2)=0\\\\\therefore \ \ \ \ (2x-1)(2x-1)(x+2)=0\end{array}$

$ \displaystyle \therefore \ \ \ \ x=\frac{1}{2}\ \text{or}\ x=-2$

5.       The remainder when the expression $ \displaystyle x^3+9x^2+bx+c$ is divided by $ \displaystyle x-2$ is twice the remainder when the expression is divided by $ \displaystyle x-1$. Show that $ \displaystyle c = 24$. Given that $ \displaystyle x+8$ is a factor of $ \displaystyle x^3+9x^2+bx+24$, show that the equation $ \displaystyle x^3+9x^2+bx+c=0$ has only one real root.

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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ f(x)={{x}^{3}}+9{{x}^{2}}+bx+c\\\\\ \ \ \ \text{By the problem},\\\\\ \ \ \ f(2)=2f(1)\\\\\ \ \ \ {{2}^{3}}+9\cdot {{2}^{2}}+2b+c=2\left( {{{1}^{3}}+9\cdot {{1}^{2}}+b+c} \right)\\\\\ \ \ \ 8+36+2b+c=2+18+2b+2c\\\\\therefore \ \ c=24\\\\\ \ \ \ \text{Since}\ x+8\ \text{is a factor of}\ {{x}^{3}}+9{{x}^{2}}+bx+24,\\\\\ \ \ \ f(-8)=0\\\\\ \ \ \ {{(-8)}^{3}}+9{{(-8)}^{2}}+b(-8)+24=0\\\\\ \ \ \ {{(-8)}^{3}}+9{{(-8)}^{2}}+b(-8)+24=0\\\\\therefore \ \ b=11\\\\\therefore \ f(x)={{x}^{3}}+9{{x}^{2}}+11x+24\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x}^{2}}+\ \ x+\ \ 3\\x+8\ \ \overline{\left){\begin{array}{l}{{x}^{3}}+9{{x}^{2}}+11x+24\\\underline{{{{x}^{3}}+8{{x}^{2}}}}\\\ \ \ \ \ \ \ \ {{x}^{2}}+11x\\\ \ \ \ \ \ \ \ \underline{{{{x}^{2}}+\ \ 8x}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3x+24\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{3x+24}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$

$ \displaystyle \therefore \ \ f(x)=(x+8)({{x}^{2}}+x+3)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =(x+8)\left[ {{{x}^{2}}+2\left( {\frac{1}{2}} \right)x+\frac{1}{4}+\frac{{11}}{4}} \right]$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ =(x+8)\left[ {{{{\left( {x+\frac{1}{2}} \right)}}^{2}}+\frac{{11}}{4}} \right]$

$ \displaystyle \ \ \ \ \text{When}\ f(x)=0,(x+8)\left[ {{{{\left( {x+\frac{1}{2}} \right)}}^{2}}+\frac{{11}}{4}} \right]=0$

$ \displaystyle \therefore \ \ x=-8\ \text{or}\ {{\left( {x+\frac{1}{2}} \right)}^{2}}=-\frac{{11}}{4}$

$ \displaystyle \ \ \ \ \text{Since}\ {{\left( {x+\frac{1}{2}} \right)}^{2}}\ge 0,{{\left( {x+\frac{1}{2}} \right)}^{2}}=-\frac{{11}}{4}\ \text{is impossible}\text{.}$

$ \displaystyle \therefore \ \ \text{There is no other real solution for }{{x}^{3}}+9{{x}^{2}}+11x+24=0.$

6.       Given that the remainder when $ \displaystyle f(x) = x^3 - x^2 + ax$ is divided by $ \displaystyle x + a$, where $ \displaystyle a > 0$, is twice the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x - 2a$, find the value of $ \displaystyle a$. Find also the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x – 2$.

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$ \displaystyle \begin{array}{l}\ \ \ \ f(x)={{x}^{3}}-{{x}^{2}}+ax\\\\\ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ f(-a)=2f(2a)\\\\\ \ \ \ -{{a}^{3}}-{{a}^{2}}-{{a}^{2}}=2\left( {8{{a}^{3}}-4{{a}^{2}}+2{{a}^{2}}} \right)\\\\\ \ \ \ 17{{a}^{3}}-2{{a}^{2}}=0\\\\\ \ \ \ {{a}^{2}}(17a-2)=0\end{array}$

$ \displaystyle \ \ \ \ \text{Since}\ a>0,17a-2=0\Rightarrow a=\frac{2}{{17}}$

$ \displaystyle \therefore \ \ f(x)={{x}^{3}}-{{x}^{2}}+\frac{2}{{17}}x$

$ \displaystyle \begin{array}{l}\ \ \ \ \text{When}\ f(x)\ \text{is divided by}\ x-2,\\\ \ \ \ \text{the remainder is}\ f(2).\end{array}$

$ \displaystyle \therefore \ \ f(2)={{(2)}^{3}}-{{(2)}^{2}}+\frac{2}{{17}}(2)=\frac{{72}}{{17}}$

7.       Given that $ \displaystyle f(x) = x^{2n} - ( p + 1)x^2 + p$, where $ \displaystyle n$ and $ \displaystyle p$ are positive intergers. Show that $ \displaystyle x - 1$ is a factor of $ \displaystyle f(x)$ for all values of $ \displaystyle p$. When $ \displaystyle p = 4$, find the value of $ \displaystyle n$ for which $ \displaystyle x - 2$ is a factor of $ \displaystyle f(x)$ and, for this case, hence factorise $ \displaystyle f(x)$ completely.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ f(x)={{x}^{{2n}}}-(p+1){{x}^{2}}+p\ \text{where}\ n,p\in {{J}^{+}}\\\\\ \ \ \ \ f(1)={{(1)}^{{2n}}}-(p+1){{(1)}^{2}}+p\\\\\ \ \ \ \ \text{Since}\ n,p\in {{J}^{+}},{{(1)}^{{2n}}}=1\ \text{and}-(p+1)=-p-1.\\\\\therefore \ \ \ f(1)=1-p-1+p=0\\\\\therefore \ \ \ x-1\ \text{is a factor of }f(x)\ \text{for all values of}\ p.\\\\\ \ \ \ \ \text{When}\ p=4,f(x)={{x}^{{2n}}}-5{{x}^{2}}+4.\\\\\ \ \ \ \ x-2\ \text{is a factor of }f(x).\\\\\therefore \ \ \ f(2)=0\\\\\ \ \ \ \ {{2}^{{2n}}}-5\cdot {{2}^{2}}+4=0\\\\\therefore \ \ \ {{2}^{{2n}}}=16\Rightarrow n=2\end{array}$

8.       The polynomial $ \displaystyle f(x)$ is given by $ \displaystyle f(x) = ax^3 + 11x^2 + cx - 60$ , where $ \displaystyle a$ and $ \displaystyle c$ are constants. If the roots of $ \displaystyle f(x) = 0$ are $ \displaystyle 2 , - 3$ and $ \displaystyle k$, find the values of $ \displaystyle a , c$ and $ \displaystyle k$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ f(x)=a{{x}^{3}}+11{{x}^{2}}+cx-60\\\\\ \ \ \ \ \text{Since the roots of}\ f(x)=0\ \text{are}\ 2,-3\ \text{and}\ k,\\\\\ \ \ \ \ a{{x}^{3}}+11{{x}^{2}}+cx-60=(x-2)(x+3)(x-k)\\\\\therefore \ \ \ a{{x}^{3}}+11{{x}^{2}}+cx-60={{x}^{3}}+(1-k){{x}^{2}}-(6+k)x+6k\\\\\ \ \ \ \ \text{Equating the corresponding terms,}\\\\\ \ \ \ \ a=1,\\\\\ \ \ \ \ 6k=-60\Rightarrow k=-10\\\\\ \ \ \ \ c=-(6+k)=-(6-10)=4\end{array}$

9.       Given that $ \displaystyle 4x^4 - 9a^2x^2 + 2(a^2 - 7) x - 18$ is exactly divisible by $ \displaystyle 2x - 3a$, show that $ \displaystyle a^3 - 7a - 6 = 0$ and hence find the possible values of $ \displaystyle a$.

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$ \displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)=4{{x}^{4}}-9{{a}^{2}}{{x}^{2}}+2({{a}^{2}}-7)x-18\\\\\ \ \ \ \ \text{Since }f(x)\ \text{is exactly divisible by}\ 2x-3a,\end{array}$

$ \displaystyle \ \ \ \ \ f\left( {\frac{{3a}}{2}} \right)=0.$

$ \displaystyle \therefore \ \ \ 4{{\left( {\frac{{3a}}{2}} \right)}^{4}}-9{{a}^{2}}{{\left( {\frac{{3a}}{2}} \right)}^{2}}+2({{a}^{2}}-7)\left( {\frac{{3a}}{2}} \right)-18=0$

$ \displaystyle \ \ \ \ \ \ 4\left( {\frac{{81{{a}^{4}}}}{{16}}} \right)-9{{a}^{2}}\left( {\frac{{9{{a}^{2}}}}{4}} \right)+3{{a}^{3}}-21a-18=0$

$ \displaystyle \begin{array}{l}\therefore \ \ \ \ {{a}^{3}}-7a-6=0\\\\\ \ \ \ \ \text{Let }g(a)={{a}^{3}}-7a-6\\\\\ \ \ \ \ g(-1)=-1+7-6=0\\\\\therefore \ \ \ \ a+1\ \text{is a factor of }g(a).\\\\\therefore \ \ \ \ \text{Let }g(a)=(a+1)({{a}^{2}}+pa+q),\\\ \ \ \ \ \ \text{where }p\ \text{and }q\ \text{are any real constants}\text{.}\\\\\therefore \ \ \ g(a)={{a}^{3}}+(p+1){{a}^{2}}+(p+q)a+q\\\\\therefore \ \ \ {{a}^{3}}-7a-6={{a}^{3}}+(p+1){{a}^{2}}+(p+q)a+q\\\\\ \ \ \ \ \text{Equating the corresponding terms,}\\\\\ \ \ \ \ p+1=0\Rightarrow p=-1\ \ q=-6\\\\\therefore \ \ \ {{a}^{3}}-7a-6=(a+1)({{a}^{2}}-a-6)\\\\\therefore \ \ \ {{a}^{3}}-7a-6=(a+1)(a+2)(a-3)\\\\\ \ \ \ \text{When }{{a}^{3}}-7a-6=0,\\\\\ \ \ \ (a+1)(a+2)(a-3)=0\\\\\therefore \ \ a=-1\ \text{or }a=-2\ \text{or }a=3.\end{array}$

10.      Solve the equation $ \displaystyle 4x^3 + 3x^2 -16x = 12$. Hence find the value of $ \displaystyle x$ such that $ \displaystyle 4e^{3x} + 3e^{2x} - 16e^x = 12$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x=12\\\\\therefore \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x-12=0\\\\\ \ \ \ \ \text{Let}\ f(x)=4{{x}^{3}}+3{{x}^{2}}-16x-12\\\\\ \ \ \ \ f(2)=4{{(2)}^{3}}+3{{(2)}^{2}}-16(2)-12\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =32+12-32-12\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =0\\\\\therefore \ \ \ (x-2)\ \text{is a factor of }f(x).\end{array}$

$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4{{x}^{2}}+11x+\ \ 6\\\ \ \ \ \ x-2\overline{\left){\begin{array}{l}4{{x}^{3}}+3{{x}^{2}}-16x-12\\\underline{{4{{x}^{3}}-8{{x}^{2}}}}\\\ \ \ \ \ \ \ \ \ \ 11{{x}^{2}}-16x\\\ \ \ \ \ \ \ \ \ \ \underline{{11{{x}^{2}}-22x}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x-12\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{{6x-12}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\end{array}}\right.}\end{array}$

$ \displaystyle \begin{array}{l}\therefore \ \ f(x)=(x-2)\left( {4{{x}^{2}}+11x+\ \ 6} \right)\\\\\therefore \ \ f(x)=(x-2)(x+2)(4x+3)\\\\\ \ \ \ 4{{x}^{3}}+3{{x}^{2}}-16x-12=0\\\\\therefore \ \ (x-2)(x+2)(4x+3)=0\end{array}$

$ \displaystyle \therefore \ \ x=2\ \text{or}\ x=-2\ \text{or}\ x=-\frac{3}{4}$

$ \displaystyle \begin{array}{l}\ \ \ \ 4{{e}^{{3x}}}+3{{e}^{{2x}}}-16{{e}^{x}}=12\\\\\ \ \ \ 4{{({{e}^{x}})}^{3}}+3{{({{e}^{x}})}^{2}}-16{{e}^{x}}-12=0\end{array}$

$ \displaystyle \therefore \ \ {{e}^{x}}=2\ \text{or}\ {{e}^{x}}=-2\ \text{or}\ {{e}^{x}}=-\frac{3}{4}$

$ \displaystyle \ \ \ \ \text{Since}\ {{e}^{x}}>0\ \text{for all }x\in R,$

$ \displaystyle \ \ \ \ {{e}^{x}}=-2\ \text{or}\ {{e}^{x}}=-\frac{3}{4}\ \text{is impossible}$

$ \displaystyle \therefore \ \ {{e}^{x}}=2\Rightarrow x=\ln 2\ $

ေမးခြန္းေဟာင္းမ်ား ႏွင့္ IGCSE O' Level မွ ေမးခြန္းမ်ား ျဖစ္ပါသည္။

Remainder Theorem : 2016 တကၠသိုလ္၀င္တန္းေမးခြန္း

The cubic polynomial $ \displaystyle f(x)$ is such that the coefficient of $ \displaystyle x^3$ is $ \displaystyle -1$ and the roots of the equation $ \displaystyle f(x)=0$ are $ \displaystyle 1$, $ \displaystyle 2$ and $ \displaystyle k$. Given that $ \displaystyle f(x)$ has a remainder of $ \displaystyle 8$ when divided by $ \displaystyle x-3$, find the value of $ \displaystyle k$ and the remainder when $ \displaystyle f(x)$ is divided by $ \displaystyle x+3$ . 

$ \displaystyle f(x)$ ဟာ အႀကီးဆံုးထပ္ကိန္း $ \displaystyle 3$ ထပ္ပါတဲ့ $ \displaystyle \text{polynomial}$ ကိန္းတန္း တစ္ခုျဖစ္တယ္။ $ \displaystyle f(x)=0$ ကို ေျဖရွင္းလို႔ရတဲ့ $ \displaystyle x$ တန္ဖိုးေတြက $ \displaystyle 1$, $ \displaystyle 2$ နဲ႔ $ \displaystyle k$ ျဖစ္ၾကတယ္။ $ \displaystyle x^3$ ရဲ့ ေျမႇာက္ေဖၚကိန္းက $ \displaystyle -1$ ျဖစ္တယ္။ 

$ \displaystyle f(x)$ ကို $ \displaystyle x-3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $ \displaystyle \text{(remainder)}$ က $ \displaystyle 8$ ျဖစ္တယ္ဆိုရင္ $ \displaystyle k$ ကို ရွာေပးပါ။

အဲဒီေနာက္ $ \displaystyle f(x)$ ကို $ \displaystyle x+3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $ \displaystyle \text{(remainder)}$ ကိုလည္း ရွာေပးပါ။

ဒီအခ်က္ေတြကိုေပါင္းလိုက္ရင္

• $ \displaystyle f(x)=-1(x-1)(x-2)(x-k)$ လို႔သိရပါမယ္။


• $ \displaystyle f(3)= 8$ ျဖစ္တယ္လို႔ သိရမယ္။


• $ \displaystyle f(3)$ မွာ ကို $ \displaystyle \text{function}$ မွာ အစားသြင္းၿပီး $ \displaystyle f(3)$ နဲ႔ ညီေပးလိုက္ရင္ $ \displaystyle k$ တန္ဖိုးရၿပီေပါ့။


• $ \displaystyle k$ ရတဲ့အခါ $ \displaystyle f(x)$ ကို $ \displaystyle x+3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $ \displaystyle \text{(remainder)}$ ဆိုတာ $ \displaystyle f(-3)$ ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။

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Solution

$ \displaystyle \begin{array}{l}\text{By the problem,}\\\\f(x)=-1(x-1)(x-2)(x-k)\\\\f(3)=8\\\\\therefore -1(3-1)(3-2)(3-k)=8\\\\\therefore -1(2)(1)(3-k)=8\\\\\therefore 3-k=-4\Rightarrow k=7\\\\\therefore f(x)=-1(x-1)(x-2)(x-7)\\\\\text{When }f(x)\ \text{is divided by }x+3,\\\\\text{the remainder = }f(-3)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-3-1)(-3-2)(-3-7)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-4)(-5)(-10)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =200\end{array}$

Problem Study: The Remainder Theorem and Division Algorithm

Given that x⁵ + ax³ + bx² − 3 = (x² − 1) Q(x) − x − 2 where Q(x) is a polynomial. State the degree of Q(x) and find the value of a and b. Find also the remainder when Q(x) is divided by x + 2. 

Solution

Since $ \displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3$ is divided by (x² − 1), Q(x) is a polynomial of degree 3.  

$ \displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$

$ \displaystyle {{x}^{5}}+a{{x}^{3}}+b{{x}^{2}}-3=(x-1)(x+1)Q(x)-x-2$

 When x = − 1, $ \displaystyle -1-a+b-3=(-1-1)(-1+1)Q(x)-(-1)-2$

$ \displaystyle \therefore -a+b=3$      --------------(1)

When x = 1, $ \displaystyle 1+a+b-3=(1-1)(1+1)Q(x)-1-2$

$ \displaystyle \therefore a+b=-1$    --------------(2)

$ \displaystyle (2)+(1)\Rightarrow 2b=2\Rightarrow b=1$

$ \displaystyle (2)-(1)\Rightarrow 2a=-4\Rightarrow a=-2$ 

$ \displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}-3=({{x}^{2}}-1)Q(x)-x-2$

$ \displaystyle \therefore {{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1=({{x}^{2}}-1)Q(x)$ 

$ \displaystyle \therefore Q(x)=\frac{{{{x}^{5}}-2{{x}^{3}}+{{x}^{2}}+x-1}}{{{{x}^{2}}-1}}$

When Q(x) is divided by x + 2, 

the remainder $ \displaystyle =Q(-2)$ 

                       $ \displaystyle =\frac{{{{{(-2)}}^{5}}-2{{{(-2)}}^{3}}+{{{(-2)}}^{2}}+(-2)-1}}{{{{{(-2)}}^{2}}-1}}$

                       $ \displaystyle =\frac{{-32+16+4-2-1}}{3}$

                       $ \displaystyle =-\frac{{15}}{3}$

                       $ \displaystyle =-5$

Problem Study (Polynomial)

    When a polynomial f(x) is divided by (x - 1) and (x + 5),
    the remainders are -6 and 6 respectively. Let r(x be the
    remainder when f(x) is divided by x² + 4x - 5. Find the  
    value of r(-2).   

Solution 

    By the problem,

    f(x) = p(x) (x - 1) - 6 ............(1)

    f(x) = q(x) (x + 5) + 6 ............(2)

    f(x) = Q(x) (x² + 4x - 5) + r(x) ............(3) 

    (1) × (x + 5) ⇒ (x + 5) f(x) = p(x) (x - 1) (x + 5) - 6x - 30   

    (2) × (x - 1)  ⇒  (x - 1) f(x) = q(x) (x - 1) (x + 5) + 6x - 6 

    Subtracting the two equations,
    6 f(x) = [p(x) - q(x)] (x - 1) (x + 5) - 12x - 24
    Hence ,
    f(x) = ${\displaystyle \frac{p\left(x\right)-q\left(x\right)}{6}}$(x - 1) (x + 5) - 2x - 4 -------(4)
    Comparing equations (3) and (4), we have
    r(x) = - 2x - 4
    Therefore, r (-2) = -2 (-2) - 4 = 0.

    Credit : Sayar U Pyi Kyaw

The Factor Theorem

6 = 3 × 2

6÷2 => remainder = 0

6÷3 => remainder = 0

အၾကြင္း 0 ရေအာင္စားႏိုင္ေသာ စားကိန္းကို တည္ကိန္း၏ factor ဟုေခၚသည္။

The Factor Theorem:

Let f(x) be a polynomial. Then (x-k) is a factor of f(x) if and only if f(k) = 0.

Example 1

Find what value p must have in order that x - p may be a factor of
4x³ - (3p + 2)x² - (p² - 1)x + 3.

Let f(x) = 4x³ - (3p + 2)x² - (p² - 1)x + 3.
x - p is a factor of f(x) only if
f(p) = 0
4p³ - (3p + 2)p² - (p² - 1)p + 3 = 0
2p² - p - 3 = 0
(p + 1) (2p - 3) = 0
p = -1 or p = 3/2