Question

How do you Math problems like 3y^2 - 15y - 252 or 2x^2 + 11x + 12 or xw - yw - xz - yz by factoring the polynomials?

Answer 1

1. How do you Math problems like $3y^{2} - 15y - 252$ or $2x^{2} + 11x + 12$ or $xw - yw - xz - yz$ by factoring the polynomials?

Explanation

First problem $3y^{2} - 15y - 252$

First we will notice that we can factor a 3 out of every term.

$3(y^{2} - 5y - 84)$

We can always check our factoring by multiplying the terms back out to make sure we get the original polynomial. Here is the form of a quadratic trinomial with argument $y \left[ (y^2 - 5y - 84) \right]$ . To solve this problem we multiplying $a$ and $c$ ( $a = 1, c = -84$ ). We get (1)(-84) = -84

We can subtract the pairs to find the differences. If there is a pair of factors with a difference of 5, then we can factor the quadratic. Now that we have factor pair (with the larger number having the "minus" sign), factor the quadratic:

$3(y^{2} - 12y + 7y - 84) = 3\big(y(y + 7) - 12(y + 7)\big) = 3((y + 7)(y - 12))$

Also we can solve a quadratic equation $y^{2} - 5y - 84$ in the form: $ay^2 + by + c$

$$y _ {1, 2} = \frac {- b \pm \sqrt {b ^ {2} - 4 a c}}{2 a} = \frac {5 \pm \sqrt {2 5 + 4 \cdot 8 4}}{2} = \frac {5 \pm 1 9}{2}$$ $$y _ {1} = 1 2, y _ {2} = - 7$$

$3y^{2} - 15y - 252 = 3((y - 12)(y + 7))$

1. Another math problem $\sqrt[2]{x^2 + 11x + 12}$ also can be solved by ac-method.

Multiplying $a$ and $c$ ( $a = 2, c = 12$ ). We get(2)(12) = 24.

Substitute the obtained values: $2x^{2} + 8x + 3x + 12$ . Apply the method of grouping

$2x^{2} + 8x + 3x + 12 = 2x(x + 4) + 3(x + 4) = (x + 4)(2x + 3)$

$$x _ {1} = - 4, x _ {2} = - \frac {3}{2}$$

Similarly, the problem can be solved by finding the roots of a quadratic equation.

2. Math problem $xw - yw - xz - yz$. Apply the method of grouping $xw - yw - xz - yz = w(x - y) - z(x + y)$ can't be factored.

$xw - yw - xz + yz$ can be factoring $xw - yw - xz + yz = w(x - y) - z(x - y) = (w - z)(x - y)$