# Question #5f8be

Sep 15, 2017

$\left(5 x - 4\right) \left(3 x + 1\right)$

#### Explanation:

I remember receiving this problem before . . . Here's how my math teacher explained it:

You can't factor this the way you normally would with quadratic equations, becuase there are no two numbers whose sum is 7 and whose product is 4.

So, you just have to experiment with the problem by inputting factors of $- 4$ into

$\left(x - a\right) \left(x + b\right)$

$\left(5 x - 4\right) \left(3 x + 1\right)$

If you FOIL the expression, you'll end up with the original equation.

Sep 15, 2017

$x$ = $\frac{4}{5}$ and - $\frac{1}{3}$

#### Explanation:

In short, bring over the 4 and factor by grouping
You always have to make the equation equal to zero when factoring so in this case you would bring the 4 over.
$15 {x}^{2}$$-$$7 x$$-$4

NOTE: you can get rid of the zero, it doesn't matter now

Then factor by grouping,
- Multiply the last number by the leading coefficient, in this case 15.
15 * -4 = -60

• Don't replace the - 4 with -60 just find two numbers that add up to -7, but multiplies to -60
• In this case it would be -12 and 5
• Now replace the -$7 x$ with -$12$ and $5$

$15 {x}^{2}$$+$$5 x$-12x$-$4

NOTE: I placed them the way they are so that when we factor it, it will give us whole numbers

• Then factor by splitting the equation in half
• Factor $15 {x}^{2}$$+$$5 x$ and $- 12 x$$-$4 individually
• Add brackets to separate them

$5 x$($3 x$$+$1) $-$ 4 (3$x$ + 1)

Now since the numbers are the same you can write it as one, then combine what's outside of the bracket together

($5 x$$-$4)($3 x$$+$1)

NOTE: You know you did right if the numbers in the brackets are the same

NORMALLY, you would be done, however the $x$'s need to be by itself, so you would solve for $x$ within each bracket by making it equal to zero

$5 x$$-$4 = 0
$5 x$ = 4
$x$ = $\frac{4}{5}$

$3 x$$+$1 = 0
$3 x$ = - 1
$x$ = -$\frac{1}{3}$

Therefore, $x$ = $\frac{4}{5}$ and - $\frac{1}{3}$

Sep 15, 2017

$- \frac{1}{3} \mathmr{and} \frac{4}{5}$

#### Explanation:

To avoid doing the lengthy factoring by grouping, you may use the new Transforming Method (Socratic, Google Search)
$y = 15 {x}^{2} - 7 x - 4 = 0$
Converted equation:
$y ' = {x}^{2} - 7 x - 60$.
Proceeding: Find 2 real roots of y', then, divide them by a = 15
The 2 real roots of y' are: - 5 and 12 -->
[Sum = 7 = - b] and Product {ac = - 60]
The 2 real roots of y are: $x 1 = - \frac{5}{15} = - \frac{1}{3}$ and $x 2 = \frac{12}{15} = \frac{4}{5}$.