# Factor polynomials on the form of ax^2 + bx +c

In the same way as you could factor trinomials on the form of

$$x^{2}+bx+c$$

You can factor polynomials on the form of

$$ax^{2}+bx+c$$

If a is positive then you just proceed in the same way as you did previously except now

$$ax^{2}+bx+c=\left ( x+m \right )\left ( ax+n \right )$$

$$where\: \: c=mn,\: \: ac=pq\: \: \\and\: \: b=p+q=am+n$$

**Example**

$$3x^{2}-2x-8$$

We can see that c (-8) is negative which means that m and n does not have the same sign. We now want to find m and n and we know that the product of m and n is -8 and the sum of m and n multiplied by a (3) is b (-2) which means that we're looking for two factors of -24 whose sum is -2 and we also know that one of them is positive and of them is negative.

$$\begin{matrix} Factors\: of\: -24 \: \: \: & Sum\: of\: factors\\ -1, 24 & 23\\ 1,-24&-23 \\ -2,12&10 \\ 2,-12&-10 \\ -3,8&5 \\ 3,-8& -5\\ -4,6&2 \\ {\color{green} {4,-6}}&{\color{green} {-2}} \end{matrix}$$

This means that:

$$3x^{2}-2x-8=$$

$$=3x^{2}+\left ( 4-6 \right )x-8 =$$

$$=3x^{2}+4x-6x-8$$

We can then group those terms that have a common monomial factor. The first two terms have x together and the second two -2 and then factor the two groups.

$$=\begin{pmatrix} 3x^{2}+4x \end{pmatrix}+\begin{pmatrix} -6x-8 \end{pmatrix}= $$

$$=x\begin{pmatrix} 3x+4 \end{pmatrix}-2\begin{pmatrix} 3x+4 \end{pmatrix} $$

Notice that both remaining parenthesis are the same. This means that we can rewrite this using the distributive propertyit as:

$$=\begin{pmatrix} x-2 \end{pmatrix}\begin{pmatrix} 3x+4 \end{pmatrix}=3x^{2}-2x-8$$

This method is called factor by grouping.

A polynomial is said to be factored completely if the polynomial is written as a product of unfactorable polynomials with integer coefficients.

**Video lesson**

Factor the following polynomial

$$2x^2+10+12$$