Question

Factorise:
$x^2-17x+60$

Answer 1

In the given polynomial $x^2-17x+60$,

The first term is $x^2$ and its coefficient is $1$.

The middle term is $-17x$ and its coefficient is $-17$.

The last term is a constant term $60$.

Multiply the coefficient of the first term by the constant $1\times 60=60$.

We now find the factor of $60$ whose sum equals the coefficient of the middle term, which is $-17$ and then factorize the polynomial $x^2-17x+60$ as shown below:

$x^2-17x+60=x^2-12x-5x+60=x(x-12)-5(x-12)=(x-5)(x-12)$

Hence, $x^2-17x+60=(x-5)(x-12)$.