# 8x^2-14x+3? I dont understand how to solve this using factoring grouping?

## $8 {x}^{2} - 14 x + 3$

Mar 8, 2017

#### Explanation:

It is important to know that when you try to factorize a quadratic polynomial say $a {x}^{2} + b x + c$ by grouping,

1. Remember you can do it, only if the determinant ${b}^{2} - 4 a c$ is a complete square of a number . This is generally not a big deal as generally in problems where factoring by grouping is desired, this is ensured. However, if you are unable to do this you can check for this. Here in $8 {x}^{2} - 14 x + 3$, we have $a = 8$, $b = - 14$ and $c = 3$. Hence ${b}^{2} - 4 a c = {\left(- 14\right)}^{2} - 4 \times 8 \times 3 = 196 - 96 = 100 = {10}^{2}$ and you can factorize the given polynomial by grouping.

2. Now you have to split $b$ in two parts, whose sum is $b$ and product is $a \times c$. Here to make things easier look at the signs of $a$ and $c$ . If they are same, look at their sum equal to $b$ and if signs are different, look at their difference equal to $b$.
Here, signs of $a = 8$ and $b = 3$ are same, as they are bot positive (an example of different signs given below), hence find two numbers whose sum is $14$ and product is $8 \times 3 = 24$.

3. For making things easier, particularly when $a \times c$ is a large number, list factors as shown here $\left(\begin{matrix}1 & 24 \\ 2 & 12 \\ 3 & 8 \\ 4 & 6\end{matrix}\right)$ and we end here as the factors are getting repeated there after. Here sum of $2$ and $12$ is equal to $14$, hence split accordingly.

Now $8 {x}^{2} - 14 x + 3$

= $8 {x}^{2} - 2 x - 12 x + 3$

= $2 x \left(4 x - 1\right) - 3 \left(4 x - 1\right)$

= $\left(2 x - 3\right) \left(4 x - 1\right)$

Had the polynomial been $8 {x}^{2} - 5 x - 3$, as signs of $a$ and $c$ are opposite, and difference is desired to be $5$, we should have chosen $3$ and$8$ and factors would have been

$8 {x}^{2} - 5 x - 3$

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

= $8 x \left(x - 1\right) + 3 \left(x - 1\right)$

= $\left(8 x + 3\right) \left(x - 1\right)$