# 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 $\displaystyle =Q(-2)$

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

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

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

$\displaystyle =-5$