# Can y= x^2-3x-28  be factored? If so what are the factors ?

##### 1 Answer
Dec 7, 2015

${x}^{2} - 3 x - 28 = \left(x - 7\right) \left(x + 4\right)$

#### Explanation:

Luckily, your ${x}^{2}$ term already has the factor $1$ in front - this makes your computation easier.

I would like to suggest the following method:

If your factorization exists, it would look like this:

${x}^{2} - 3 x - 28 = \left(x + a\right) \left(x + b\right)$

$\textcolor{w h i t e}{\times \times \times \times x} = {x}^{2} + \left(a + b\right) x + a \cdot b$

So, your goal is to find $a$ and $b$ with the following properties:

• $a + b = - 3$
• $\textcolor{w h i t e}{i} a \times b \textcolor{w h i t e}{i} = - 28$

It is always easier to start with the multiplication. Since $- 28$ is negative, one of the two numbers needs to be positive and the other one negative.
Possible integer pairs of $a$ and $b$ are:

• $\left(- 4\right) \times 7 \textcolor{w h i t e}{\times}$ or $\textcolor{w h i t e}{\times} 4 \times \left(- 7\right)$
• $\left(- 2\right) \times 14 \textcolor{w h i t e}{\xi i}$ or $\textcolor{w h i t e}{\times} 2 \times \left(- 14\right)$
• $\left(- 1\right) \times 28 \textcolor{w h i t e}{\xi i}$ or $\textcolor{w h i t e}{\times} 1 \times \left(- 28\right)$

It is easy to check that the correct pair is $4$ and $- 7$ since $4 + \left(- 7\right) = - 3$ and $4 \cdot \left(- 7\right) = - 28$.

Thus, your factorization is

${x}^{2} - 3 x - 28 = \left(x - 7\right) \left(x + 4\right)$.

$\textcolor{w h i t e}{x}$

By the way, this also means that the solutions for the equation

${x}^{2} - 3 x - 28 = 0$

are $- a$ and $- b$, so $x = 7$ or $x = - 4$.