Question

Why are $\frac{1}{4}$ and $-1$ zeroes of the polynomial $p\left(x\right)=4x^{2}x+3x-1$?

Answer 1

Polynomial $p\left(x\right)=4x^{2}x+3x-1$
The zeroes of its are $\frac{1}{4}and-1$
Let $x=\frac{1}{4}$
$p\left(\frac{1}{4}\right)=4{\left(\frac{1}{4}\right)}^2+3\left(\frac{1}{4}\right)-1$
$=4\times \frac{1}{16}+\frac{3}{4}-1$
$=\frac{1}{4}+\frac{3}{4}-1$
$=\frac{1+3-4}{4}=\frac{0}{4}=0$
Let $P\left(x\right)=-1$
$P(-1)=4{(-1)}^2+3(-1)-1$
$=4-3-1=0$
$\therefore p\left(\frac{1}{4}\right)=0$ and $P(-1)=0$
Since $\frac{1}{4}$ and $-1$ are zeroes of
The polynomial $p\left(x\right)$.