Non-monic quadratic trinomials (a not 1)
Lesson

So far, most of the quadratics we've dealt with are monic, meaning their $x^2$x2 term only has a coefficient of $1$1. If the coefficient is not $1$1, then we've usually found we can factor out that coefficient from the whole quadratic.

eg. $2x^2-4x+6=2\left(x^2-2x+3\right)$2x24x+6=2(x22x+3).

But how do we factor quadratics that can't be simplified in this way? First let's have a look at how a non-monic quadratic is composed:

Now we are more familiar with these tricky quadratics let's have a look at the three different methods below.

## Cross Method

We've already encountered the cross method once before with monic quadratics, and it's easy to see how this extends into non-monic territory.

For example, let's have a look at $5x^2+11x-12$5x2+11x12. We must draw a cross with a possible pair of factors of $5x^2$5x2 on one side and another possible factor pair of $-12$12 on the other side.

Let's start with the factor pairs of $5x$5x & $x$x on the left, and $-6$6 & $2$2 on the other:

$5x\times2+x\times\left(-6\right)=4x$5x×2+x×(6)=4x, which is incorrect, so let's try again with another two pairs:

$5x\times3+x\times\left(-4\right)=11x$5x×3+x×(4)=11x which is the right answer. By reading across in the two circles, the quadratic must then factor to $\left(5x-4\right)\left(x+3\right)$(5x4)(x+3).

## Factor By Grouping Method

The PSF (Product, Sum, Factor) method uses a similar idea we had with monic quadratics where we think about sums and products, but slightly different.

Procedure

For a quadratic in the form $ax^2+bx+c$ax2+bx+c:

1. Find two numbers, $m$m & $n$n, that have a SUM of $b$b and a PRODUCT of $ac$ac.

2. Rewrite the quadratic as $ax^2+mx+nx+c$ax2+mx+nx+c.

3. Use grouping in pairs to factor the four-termed expression.

#### Example

##### question 1

Using the same example as above, factor $5x^2+11x-12$5x2+11x12 using the Factor By Grouping Method.

Think about what the sum and product of $m$m & $n$n should be

Do

We want the sum of of $m$m & $n$n to be $11$11, and the product to be $5\times\left(-12\right)=-60$5×(12)=60

The two numbers work out to be $4$4 & $-15$15, so:

 $5x^2+11x-12$5x2+11x−12 $=$= $5x^2-4x+15x-12$5x2−4x+15x−12 $=$= $x\left(5x-4\right)+3\left(5x-4\right)$x(5x−4)+3(5x−4) $=$= $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3)

This is the same answer that we got before!

## PSF Variation

The above two methods are the most often used. However, a slightly different method can also be used to factor directly if you can remember the formula.

Formula

$ax^2+bx+c=\frac{\left(ax+m\right)\left(ax+n\right)}{a}$ax2+bx+c=(ax+m)(ax+n)a, where $m+n=b$m+n=b & $mn=ac$mn=ac

#### Example

##### question 2

Factor $5x^2-36x+7$5x236x+7 completely

Think about whether it is easier to consider the product or the sum of $m$m & $n$n first

Do

 $m+n$m+n $=$= $b$b $=$= $-36$−36 $mn$mn $=$= $ac$ac $=$= $5\times7$5×7 $=$= $35$35

It's much easier to look at the product first as there're less possible pairs that multiply to give $35$35 than those that add to give $-36$36. We can easily see that $m$m & $n$n $=$= $-1$1 & $-35$35. Then:

 $5x^2-36x+7$5x2−36x+7 $=$= $\frac{\left(5x-1\right)\left(5x-35\right)}{5}$(5x−1)(5x−35)5​ $=$= $\frac{\left(5x-1\right)\left(x-7\right)\times5}{5}$(5x−1)(x−7)×55​ $=$= $\left(5x-1\right)\left(x-7\right)$(5x−1)(x−7)

#### Worked Examples

##### Question 3

Factor the trinomial:

$7x^2-75x+50$7x275x+50

##### Question 4

Factor the following trinomial:

$6x^2+13x+6$6x2+13x+6

##### Question 5

Factor $-12x^2-7x+12$12x27x+12.

### Outcomes

#### 10D.QR3.02

Factor polynomial expressions involving common factors, trinomials, and differences of squares, using a variety of tools and strategies