Question

Find the zeroes of the quadratic polynomial $x^2+14x+48$ and verify them

Answer 1

The given polynomial is $x^2+14x+48$
By factorising the quadratic polynomial we get, $x^2+14x+48=(x+8)(x+6)$
The value of $x^2+14x+48$ is zero, When $x+8=0$ or $x+6=0$.
That is, when $x=-8$ or $x=-6$
The zeroes of the polynomial $x^2+14x+48$ are $-8$ and $-6$.
Let us verify the results by substituting the values.
$p\left(x\right)=x^2+14x+48=(-8{)}^2+14(-8)+48=64-112+48=0\therefore p(-8)=0$
$p(-6)=(-6{)}^2+14(-6)+48=36-84+48\therefore p(-6)=0$