Question #5f8be

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Question #5f8be

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Answer 1 · 2017-09-15T01:38:39

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

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 $- 4$ $-4$ into

$(x - a) (x + b)$ $(x-a)(x+b)$

The answer is:

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

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

Answer 2 · 2017-09-15T01:51:21

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

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.
$15 x^{2}$ $15x^2$ $-$ $-$ $7 x$ $7x$ $-$ $-$ 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 - $7 x$ $7x$ with - $12$ $12$ and $5$ $5$

$15 x^{2}$ $15x^2$ $+$ $+$ $5 x$ $5x$ -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 $15 x^{2}$ $15x^2$ $+$ $+$ $5 x$ $5x$ and $- 12 x$ $-12x$ $-$ $-$ 4 individually
Add brackets to separate them

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

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

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

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

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

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

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

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

Answer 3 · 2017-09-15T03:48:26

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

Explanation:

To avoid doing the lengthy factoring by grouping, you may use the new Transforming Method (Socratic, Google Search)
$y = 15 x^{2} - 7 x - 4 = 0$ $y = 15x^2 - 7x - 4 = 0$
Converted equation:
$y' = x^2 - 7x - 60$ .
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: $x1 = - 5/15 = -1/3$ and $x2 = 12/15 = 4/5$ .

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