# Question 5acbe

Mar 27, 2017

Although you have not formatted the question properly I think it is
C. I suspect you meant C to be $- \frac{1}{3} \mathmr{and} + 5$

#### Explanation:

Sometimes these are easier to spot than other times. As you practice more and more you will start to spot them more easily. If not then switch to the formula.

The constant of 5 is a prime number so we must have the structure:

(?x+-1)(?x+-5)

I have use the $\pm$ as I have not as yet decided which they should be.

the coefficient of ${x}^{2}$ is 3 which is also a prime number. So we must something like:

$\left(3 x \pm 1\right) \left(x \pm 5\right)$

It could be that the three needs to be in the other bracket. I have not yet tested it.

Notice that we have $- 14 x$ which is negative so the greater of the constants need to be negative giving:

$\left(3 x + 1\right) \left(x - 5\right)$ Ok! Lets test it

color(green)(color(blue)((3x+1))(x-5)

Multiply everything in the right brackets by everything in the left brackets.

$\textcolor{g r e e n}{\textcolor{b l u e}{3 x} \left(x - 5\right) \textcolor{b l u e}{\text{ "+" } 1} \left(x - 5\right)}$

$3 {x}^{2} - 15 x \text{ "+" } x - 5$

$3 {x}^{2} - 14 x - 5 \leftarrow \textcolor{red}{\text{ Works so ok!}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So we have:

$\textcolor{b l u e}{\left(3 x + 1\right)} \textcolor{g r e e n}{\left(x - 5\right)} = 0$

color(green)(x-5=0 => x=+5

color(blue)(3x+1=0 => 3x=-1=>x=-1/3#