Question

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

• A. $a=-1,b=-1;a=1,b=6$
• B. $a=3,b=0;a=2,b=3$
• C. $a=,b=-1;a=3,b=-2$
• D. $a=-1,b=2;a=1,b=2$

Answer 1

Answer: (D)

$a=-1,b=2;a=1,b=2$

$p\left(x\right)=x^5-x^4-4x^2+3x^3+3$

$q\left(x\right)=x^3+2x^2+a$

Zeroes of q(x) are also the zeroes of p(x)

Þ q(x) is a factor of p(x)

To find the another factor of p(x), it should be divided by q(x)

$x^3+2x^2+a÷x^5-x^4-4x^2+3x^3+3\setminus x^2-3x+2$

$\pm x^5\pm 2x^4\pm a{x}^2$

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$-3x^4-4x^3+(3-a)x^2+3x+b$

$\mp 3x^4\mp 6x^3\mp ax$

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$2x^3+(3-a)x^2+(3-3a)x+b$

$\pm 2x^3\pm 4x^2\pm 2a$

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$(3-a-4)x^2+(3-3a)x+(b-2a)$

$x^3-3x+2=(x-2)(x-1)$

Therefore, 2 and 1 are the other zeroes of $p\left(x\right)$.

Similarly, when p(x) is divided by $x^2-3x+2$ the quotient is $x^3-3x+2$ and remainder is $(b+2)$

$x^3-2x^2-1=x^3+2x^2+a$

$\Rightarrow a=-1$ and reminder is $0$.

$b+2=0$ then $b=-2$

Hence, the velue of $a$ and $b$ is $-1,2$ and $1,2$.