Question

Question 6
For which values of a and b, the zeroes of $q\left(x\right)=x^3+2x^2+a$ are also the zeroes of the polynomial $p\left(x\right)=x^5–x^4–4x^3+3x^2+3x+b?$ which zeroes of p (x) are not the zeroes of q(x)?

Answer 1

Given that the zeroes of $q\left(x\right)=x^3+2x^2+a$ are also the zeroes of the polynomial $p\left(x\right)=x^5-x^4-4x^3+3x^2+3x+b$ i.e., q(x) is a factor of p(x). then, we use a division algorithm
$\Rightarrow x^3+2x^2+a\stackrel{x^2-3x+2}{\sqrt{x^5-x^4-4x^3+3x^2+3x+b}}$
$\Rightarrow \frac{x^5+2x^4+a{x}^2}{-3x^4-4x^3+(3-a)x^2+3x+b}$
$\Rightarrow \frac{-3x^4-6x^3-3ax}{2x^3+(3-a)x^2+(3+3a)x+b}$
$\Rightarrow \frac{2x^3+4x^2+2a}{-(1+a)x^2+(3+3a)x+(b-2a)}$

if $(x^3+2x^2+a)$ is a factor of $(x^5-x^4-4x^3+3x^2+3x+b)$, then remainder should be zero.........$\left(Factortheorem\right)$
i.e., $-(1+a)x^2+(3+3a)x+(b-2a)=0$
$=0.x^2+0.x+0$
On comparing the coefficient of x, we get
$a+1=0$
$\Rightarrow a=-1$
and $b–2a=0$
$\Rightarrow b=2a$
$\Rightarrow b=2(-1)=-2$
For a = -1 and b = -2, the zeroes of q(x) are also the zeroes of the polynomial p(x)
$\therefore$ $1\left(x\right)=x^3+2x^2–1$
and $p\left(x\right)=x^5-5x^4-4x^3+3x^2+3x-2$
now, dividend = divisor $\times$ quotient + remainder
$p\left(x\right)=9x^3+2x^2-1\left)\right(x^2-3x+2)+0$
$=(x^3+2x^2-1)(x^2–2x–x+2)$
$(x^3+2x^2-1)(x-2)(x–1)$
Hence, the zeroes of p(x) are 1 and 2 which are not the zeroes of q (x).