Question

Write polynomial $P$ as a product of linear factors: $P\left(x\right)=x^4+4x^3+6x^2+4x-15$
[hint: $1$ and $-3$ are zeros of $P$]

Answer 1

$p\left(x\right)=x^4+4x^3+6x^2+4x-15$

Given as hint : $1\&-3$ are zeroes of $p\left(x\right)$

$\therefore$ To find other two factors we divide $p\left(x\right)$ by $(x-1)(x+3)$

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

Now, factor $x^2=2x+5$ using quadratic formula

$\Rightarrow x^2+2x+5=0$

$\Rightarrow x=\frac{-2\pm \sqrt{{\left(2\right)}^2+4\left(1\right)\left(5\right)}}{2}=\frac{-2\pm \sqrt{4-20}}{2}$

$=\frac{-2\pm 4i}{2}$

$=-1\pm 2i$

$\Rightarrow -1-2i,-1+2i\left[zero\right]$

$\therefore$ Factors are $(x+1+2i)(x+1-2i)$

So, the polynomial $p\left(x\right)$ can be written as

$(x-1)(x+3)(x+1+2i)(x+1-2i).$

Hence, solved.