Question

Q 1 ) If the polynomial $a{z}^3+4z^2+3z-4$ and $z^3-4z+a$ leave the same remainder when divided by $z-3$, find the value of $a$
Q 2) If both $x-2$ and $x-\frac{1}{2}$ are the factors of $p{x}^2+5x+r$, show that $p=r$
Q 3) Without actually division prove that $2x^4-5x^3+2x^2-x+2$ is divisible by $x^2-3x+2$

Answer 1

$1)$ Given that $a{z}^3+4z^2+3z-4$ and $z^3-4z+a$ leaves the remainder when divided by $z-3$.

Hence, $a{\left(3\right)}^3+4{\left(3\right)}^2+3\left(3\right)-4={\left(3\right)}^3-4\left(3\right)+a$

$\Rightarrow 27a+36+9-4=27-12+a$

$\Rightarrow 27a+41=15+a$

$\Rightarrow 26a=-26$

Hence, $a=-1$

$2)$By Factor theorem, we have

$(x-2)(x-\frac{1}{2})=p{x}^2+5x+r$

Consider $(x-2)(x-\frac{1}{2})=x^2-\frac{1}{2}x-2x+1$

$=x^2-\frac{5}{2}x+1$

$=-2x^2+5x-2$ on multiplying by $-2$

Now, by equating $-2x^2+5x-2=p{x}^2+5x+r$

we get $p=-2$ and $r=-2$

$\therefore p=r=-2$

$3)$Factors of $x^2-3x+2=x^2-x-2x+2$ by splitting the factors

$=x(x-1)-2(x-1)$ by taking common

$=(x-2)(x-1)$ are the factors.

$\therefore x=1,2$

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

$f\left(1\right)=2\times {\left(1\right)}^4-5\times {\left(1\right)}^3+2\times {\left(1\right)}^2-1+2=0$

$f\left(2\right)=2\times {\left(2\right)}^4-5\times {\left(2\right)}^3+2\times {\left(2\right)}^2-2+2=32-32=0$

Hence, $f\left(2\right)=0$

Hence $2x^4-5x^3+2x^2-x+2$ is divisible by $(x-2)$

$\therefore f\left(1\right).f\left(2\right)=0$

Hence, $2x^4-5x^3+2x^2-x+2$ is divisible by $x^2-3x+2$