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Factoring Harder Quadratics, where a isn’t 1

Precalculus Algebra Factoring Quadratic Expressions Factoring Harder Quadratics, where a isn’t 1

The Method

Here, we focus our attention on factoring quadratic expressions of the form

$$\begin{align}ax^2+bx+c\end{align}$$

where the leading coefficient is $a\neq 1$.

How to Factor Quadratics where $a\neq 1$.

Starting with an expression of the form $ax^2+bx+c$

1. Find two numbers that add together to get $b$ and also multiply together to get $a\cdot c$. Call them $n$ and $m$.

2. Use $m$ and $n$ to “break up the middle term” so that
$$\begin{align} ax^2+bx+c=ax^2+nx+mx+c\end{align}$$

3. Factor the expression by grouping the first pair of terms and the last pair.

4. After factoring by grouping, you should end up with an expression of the form
$$\begin{align}(\ \ \ + \ \ \ )(\ \ \ +\ \ \ )\end{align}$$

Critical Note!

The only thing that differs in this method when compared with the method for factoring quadratics where $a=1$ (i.e. where there is no number written out in front of $x^2$) is that we are looking for two numbers $n$ and $m$ that add together to give you $b$ and multiply to give you $ac$, not just $c$ as before! Nothing else in the process changes!

Technically, the process described above handles the specific cases where $a=1$ as well (because $a\cdot c=1\cdot c=c$).

Following the process above, we need to find two numbers $n$ and $m$ that when added together give you $b=7$ and when multiplied together give you $ac=2\cdot 6=12$.

Such numbers are $n=4$ and $m=3$ (since $4\cdot 3 =12=ac$ and $4+3=7=b$).

Now break apart the middle term using these numbers and then factor by grouping.

$$\begin{align}2x^2+7x+6&=2x^2+7x+3x+6\\&= (2x^2+4x)+(3x+6)\\&=2x(x+2)+3(x+2)\\&=(2x+3)(x+2)\end{align}$$

IMPORTANT NOTE: Some quadratic expressions cannot be factored in this way. You will be able to tell this by step 2, where you will be unable to find two factors of $ac$ that add to give you $b$.

$(x+1)(3x-2)$

$(x+3)(2x+1)$

$(4x+7)(x+1)$

$(4x-3)(x-5)$

$(2x-5)(3x-7)$

$(5(a+b)+1)(3(a+b)+1)$

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Precalculus Algebra

Factoring Harder Quadratics, where a isn’t 1

The Method

How to Factor Quadratics where \(a\neq 1\).

Critical Note!