Question

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

Answer 1

Let $p\left(x\right)=2x^4–5x^3+2x^2–x+2$
Factorise $x^2–3x+2$.
Now, $x^2–3x+2=x^2–2x–x+2$ [By splitting middle term]
$=x(x–2)–1(x–2)=(x–1)(x–2)$
Hence, zeros of $x^2–3x+2$ are 1 and 2.
We have to prove that, $2x^4–5x^3+2x^2–x+2$ is divisible by $x^2–3x+2$.
i.e. prove that $p\left(1\right)=0$ and $p\left(2\right)=0$
Now. $p\left(1\right)=2(1{)}^4–5(1{)}^3+2(1{)}^2–1+2$
$=2–5+2–1+2=6–6=0$
$p\left(2\right)=2(2{)}^4–5(2{)}^3+2(2{)}^2–2+2$
$=2\times 16–5\times 8+2\times 4+0$
$=32-40+8=40–40=0$
Hence, p(x) is divisible by $x^2–3x+2$.