# Factor: 7x^2 + 39x + 20

## Question:

Factor: {eq}7x^2 + 39x + 20 {/eq}

Having a trinomial in a form of {eq}f(x) = ax^2 + bx + c {/eq}, we can factor it by finding the roots of it firstly.

The trinomial is set to 0 and its roots are found applying the quadratic formula:

{eq}x_{1, 2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} {/eq}.

Having the two roots, the factored expression would have a form of {eq}a(x - x_1)(x - x_2) {/eq}.

Firstly, find the roots of the given polynomial:

{eq}7x^2 + 39x + 20 = 0...(1) {/eq}

Given a = 7, b = 39 and c = 20, we may apply the quadratic formula to find the two roots:

{eq}x_{1, 2} = \frac{-39 \pm \sqrt{39^2 - 4\cdot 7\cdot 20}}{2\cdot 7} = \frac{-39 \pm 31}{14} {/eq}

The first root is then:

{eq}\frac{-39 + 31}{14} = -\frac{8}{14} = -\frac{4}{7} {/eq}

and the second root is:

{eq}\frac{-39 - 31}{14} = -\frac{70}{14} = -5 {/eq}.

According to the formula, the factored form would then be:

{eq}7(x + \frac{4}{7})(x + 5) {/eq}.

To make the expression look neater, multiply the leading coefficient, 7, by the first term:

{eq}7(x + \frac{4}{7})(x + 5) = (7x + 4)(x + 5) {/eq}.

Thus, the factored form is (7x + 4)(x + 5).