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$