Question

If $1$ and $3$ are the zeros of the polynomial $p\left(x\right)=2x^4-7x^3-13x^2+63x-45$,then find the remaining zeros of $p\left(x\right)$

Answer 1

The given polynomial is,

$p\left(x\right)=2x^4-7x^3-13x^2+63x-45$,

the two zeros of $p\left(x\right)$ are given as 1 and 3, then $x-1$ and $x-3$ will be two factors of $p\left(x\right)$.

then, $(x-1)(x-3)=x^2-4x+3$

Now, dividing $p\left(x\right)$ by the factor $x^2-4x+3$ we can write $p\left(x\right)$ as,

$p\left(x\right)=(x^2-4x+3)(2x^2+x-15)$

We can get the other two zeros of $p\left(x\right)$ from,

$2x^2+x-15=2x^2+6x-5x-15$

$=(2x-5)(x+3)$

$\Rightarrow x=\frac{5}{2},-3$

Hence, the other two zeros of $p\left(x\right)$ are $\frac{5}{2},-3$.