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
$ \displaystyle \therefore -a+b=3$ --------------(1)
$ \displaystyle \therefore a+b=-1$ --------------(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 =\frac{{-32+16+4-2-1}}{3}$
$ \displaystyle =-\frac{{15}}{3}$
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$
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