\(\displaystyle 4x^2-28 x = 0\)\(\displaystyle 4x(x-7) = 0\)\(\displaystyle x (x-7) = 0\)\(\displaystyle \fn_cm x-7 = 0 \;or\; x = 0\)

\(\displaystyle First\;equation:x-7 = 0\)

\(\displaystyle x=7\)

\(\displaystyle x=0,7\)

$\displaystyle Find \;two \; numbers \;that \;multiply \;to \; get \;42 \; and \;add \;to \;get \; 13.$

$\displaystyle 6\times7=42$

$\displaystyle 6+7=13$

$\displaystyle Find\; two\; numbers\; that \; multiply \; to \; get\; -14 \; and\; add \; to\; get \; -5.$

$\displaystyle -7\times2=-14$

$\displaystyle -7+2=-5$

First, take out the greatest common factor of the terms. The GCF of 5 and 50 is 5 and the GCF of $\displaystyle x^{2}$ and $\displaystyle x$ is $\displaystyle x$, so the GCF of the terms is $\displaystyle 5x$.

When $\displaystyle 5x$ is distributed out, this leaves $\displaystyle 5x (x -10)$.

$\displaystyle x-10$ is linear and thus prime, so no further factoring can be done.

This polynomial is a difference of two squares. The below formula can be used for factoring the difference of any two squares.

$\displaystyle a^2-b^2=(a-b)(a+b)$

Using our given equation as $\displaystyle \small a^2-b^2$, we can find the values to use in our factoring.

$\displaystyle a^2-b^2=16x^2y^4-9w^4z^6$

$\displaystyle (4xy^2)^2=16x^2y^4\rightarrow 4xy^2=a$

$\displaystyle (3w^2z^3)^2=9w^4z^6\rightarrow 3w^2z^3=b$

$\displaystyle (a-b)(a+b)=(4xy^2-3w^2z^3)(4xy^2+3w^2z^3)$

When factoring a polynomial that has no coefficient in front of the $\displaystyle x^{2}$ term, you begin by looking at the last term of the polynomial, which is $\displaystyle -35$.  You then think of all the factors of $\displaystyle -35$ that when added together equal $\displaystyle 2$, the coefficient in front of the $\displaystyle x$ term.  The only combination of factors of $\displaystyle -35$ that can satisfy this condition is $\displaystyle -5$ and $\displaystyle 7$.  Thus, the factors of the polynomial are $\displaystyle (x-5)(x+7)$.

\(\displaystyle Factor \;x^2-4 x-32\)\(\displaystyle Find \; two \; numbers \; whose \; product \; is \; -32 \; and \; sum \; is -4.\)\(\displaystyle \fn_cm Those \; numbers \; are \; 4 \; and \; -8.\)
\(\displaystyle Therefore, \; x^2-4 x-32 = (x+4)(x-8)\)
\(\displaystyle Answer: \;(x+4)(x-8)\)

$\displaystyle x^2-3x+2$

The factored form of this equation should be in the format $\displaystyle (A+B)(C+D)$.

To yield the first term in our original equation ($\displaystyle \small x^2$), $\displaystyle \small A=x$ and $\displaystyle \small C=x$.

$\displaystyle (x+B)(x+D)$

To yield the last term in our original equation ($\displaystyle \small +2$), we can set $\displaystyle \small B=-2$ and $\displaystyle \small D=-1$.

$\displaystyle -2*-1=2\ \text{and}\ -2+(-1)=-3$

$\displaystyle (x-2)(x-1)$

We can check our answer by using FOIL to expand back to the original expression.

$\displaystyle 3x^2-2x-5$

The factored form of this equation should have the format $\displaystyle (A+B)(C+D)$.

To yield the first term in our original equation ($\displaystyle \small 3x^2$), $\displaystyle \small A=3x$ and $\displaystyle \small C=x$.

$\displaystyle (3x+B)(x+D)$

To yield the final term in our original equation ($\displaystyle \small -5$), we can set $\displaystyle \small B=-5$ and $\displaystyle \small D=1$.

$\displaystyle -5*1=-5\ \text{and}\ -5+1(3)=-2$

$\displaystyle (3x-5)(x+1)$

If you are unsure of your answer, you can check using FOIL to end up with the original equation.

$\displaystyle x^3+7x^2+3x+21$

To factor an expression like this, we can use factoring by grouping. The greatest common factor in the first two terms is $\displaystyle \small x^2$, so $\displaystyle \small x^2$ should be factored out.

$\displaystyle x^2(x+7)+3x+21$

Now, look at the last two terms. The greatest common factor in these terms is $\displaystyle \small 3$. Factor out the $\displaystyle \small 3$.

$\displaystyle x^2(x+7)+3(x+7)$

Now, both terms in parentheses are $\displaystyle \small x+7$. We can group $\displaystyle \small x^2$ and $\displaystyle \small 3$ together, and multiply it by $\displaystyle \small x+7$.

$\displaystyle (x+7)(x^2+3)$

We can check the answer using FOIL to end up with the original expression.