Question

Without actual division, prove that $2x^4-5x^3+2x^2-x+2$ is divisible by $x^2-3x+2$.

Answer 1

Step 1: Factorize $x^2-3x+2$

$\Rightarrow$$x^2-3x+2$$=$$x^2-x-2x+2$

$=x\left(x-1\right)-2\left(x-1\right)$

$\Rightarrow$$x^2-3x+2$$=\left(x-1\right)\left(x-2\right)$

Step 2: Equate $x^2-3x+2$ with $0$ to find its roots.

$\Rightarrow$$\left(x-1\right)\left(x-2\right)=0$

$\Rightarrow$$x=1$ and $x=2$

Step 3: Prove that $x^2-3x+2$ is factor of $2x^4-5x^3+2x^2-x+2$

If $x^2-3x+2$ is a factor to $P\left(x\right)=$$2x^4-5x^3+2x^2-x+2$, then roots of $x^2-3x+2$ must be roots of $P\left(x\right)$

$\Rightarrow$$P\left(1\right)=P\left(2\right)$ must be zero for $x^2-3x+2$ to be a factor of $P\left(x\right)=$$2x^4-5x^3+2x^2-x+2$

$\Rightarrow$$P\left(1\right)=2{\left(1\right)}^4-5{\left(1\right)}^3+2{\left(1\right)}^2-1+2$

$=2-5+2-1+2$

$\Rightarrow$$P\left(1\right)=0$

$\Rightarrow$$P\left(2\right)=2{\left(2\right)}^4-5{\left(2\right)}^3+2{\left(2\right)}^2-2+2$

$=32-40+8-2+2$

$\Rightarrow$$P\left(2\right)=0$

$\Rightarrow$ $x=1$ and $x=2$ are roots of the equation $P\left(x\right)=0$

Therefore, $\left(x-1\right)\left(x-2\right)$ is a factor of $P\left(x\right)=$$2x^4-5x^3+2x^2-x+2$

Hence, $\mathbf{2}{\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{5}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{2}$ is divisible by ${\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}$.