Factor:

\(\displaystyle 0=x^2+4x-12\)

\(\displaystyle \small \small 0=(x-2)(x+6)\)

Double check by factoring:

\(\displaystyle \small F: (x)(x)\)

\(\displaystyle \small \small O:(x)(6)=6x\)

\(\displaystyle \small \small I: (-2)(x)=-2x\)

\(\displaystyle \small \small L: (-2)(6)=-12\)

Add together: \(\displaystyle \small x^2+6x-2x-12=x^2+4x-12\)

Therefore:

\(\displaystyle \small x-2=0\ or x+6=0\)

\(\displaystyle \small x=2\or-6\)

1) Split up the middle term so that factoring by grouping is possible.

Factors of 10 include:

1 * 10= 10    1 + 10 = 11

2 * 5 =10      2 + 5 = 7

–2 * –5 = 10    –2 + –5 = –7 Good!

\(\displaystyle x^{2}-2x-5x+10=0\)

2) Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

\(\displaystyle x(x-2)-5(x-2)=0\)

3) Now pull out the common factor, the "(x-2)," from both terms.

\(\displaystyle (x-5)(x-2)=0\)

4) Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0,  x = 5

x – 2 = 0, x = 2

To solve for \(\displaystyle x\), you need to isolate it to one side of the equation. You can subtract the \(\displaystyle 6x\) from the right to the left. Then you can add the 6 from the right to the left:

\(\displaystyle x^2+2 = 6x-6\)

\(\displaystyle x^2-6x+2 = -6\)

\(\displaystyle x^2-6x+8 = 0\)

Next, you can factor out this quadratic equation to solve for \(\displaystyle x\). You need to determine which factors of 8 add up to negative 6:

\(\displaystyle (x \ \ \ \ \ \)(x \ \ \ \ \ \) = 0\)

\(\displaystyle (x-2)(x-4)=0\)

Finally, you set each binomial equal to 0 and solve for \(\displaystyle x\):

\(\displaystyle x=2 \ \ \ \ \ \ x=4\)

First, we have to know that "finding the roots" means "finding the values of x which make the expression =0." So basically we are going to set the original expression = 0 and factor.

\(\displaystyle 6x^2 + 5x - 4 = 0\)

This quadratic looks messy to factor by sight, so we'll use factoring by composition. We multiply a and c together, and look for factors that add to b.

\(\displaystyle 6 * -4 = -24\)

\(\displaystyle -24 = 8 * -3\)

\(\displaystyle 8 + -3 = 5\)

So we can use 8 and -3. We will re-write 5x using these numbers as 8x - 3x, and then factor by grouping.

\(\displaystyle 6x^2 + 5x - 4 = 0\)

\(\displaystyle 6x^2 + 8x - 3x - 4 = 0\)

\(\displaystyle (6x^2 + 8x) + (- 3x - 4) = 0\)

Note the extra + sign we inserted to make sure the meaning is not lost when parentheses are added. Now we identify common factors to be "pulled" out.

\(\displaystyle (3x(2x) + 4(2x)) + ((-1) 3x +(-1) 4) = 0\)

\(\displaystyle 2x(3x+4)) + -1( 3x + 4) = 0\)

Now we factor out the (3x + 4).

\(\displaystyle (3x+4)(2x-1)=0\)

Setting each factor = 0 we can find the solutions.

\(\displaystyle 2x - 1 =0\)

\(\displaystyle 2x = 1\)

\(\displaystyle x = 1/2\)

\(\displaystyle 3x + 4 = 0\)

\(\displaystyle 3x = -4\)

\(\displaystyle x = \frac{-4}{3}\)

So the solutions are x = 1/2 and x = -4/3, or {-4/3, 1/2}.

If we recognize this as an expression with form \(\displaystyle \small a^2 - b^2\), with \(\displaystyle \small a=3x\) and \(\displaystyle \small b=121\), we can solve this equation by factoring:

\(\displaystyle \small 9x^2-121=0\)

\(\displaystyle \small (3x+11)(3x-11)=0\)

\(\displaystyle \small 3x+11=0\) and \(\displaystyle \small 3x-11=0\)

\(\displaystyle \small x=-\frac{11}{3}\) and \(\displaystyle \small x=\frac{11}{3}\)

When finding zeros of a polynomial, you must remember your rules. Without a function this may seem tricky, but remember that non-real solutions come in conjugate pairs. Conjugate pairs differ in the middle sign. Thus, our answer is:

\(\displaystyle 3-i\)

\(\displaystyle \begin{align*}&\text{The function }f(x)=200x - 320x^{2} + 250\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-10\cdot (8x + 5)\cdot (4x - 5)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(200)\pm\sqrt{(200)^2-4(-320)(250)}}{2(-320)}=\frac{-200\pm600}{-640}\\&\text{Either way, we find thatthe function crosses the x-axis at} x=\frac{5}{4}\text{ and }-\frac{5}{8}.\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }f(x)=732x - 96x^{2} - 702\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&-6\cdot (8x - 9)\cdot (2x - 13)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(732)\pm\sqrt{(732)^2-4(-96)(-702)}}{2(-96)}=\frac{-732\pm516}{-192}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=\frac{13}{2}\text{ and }\frac{9}{8}.\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }f(x)=38x + 208x^{2} - 126\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&2\cdot (8x + 7)\cdot (13x - 9)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(38)\pm\sqrt{(38)^2-4(208)(-126)}}{2(208)}=\frac{-38\pm326}{416}\\&\text{Either way, we find thatthe function crosses the x-axis at} \ x=-\frac{7}{8}\text{ and }\frac{9}{13}.\end{align*}\)

\(\displaystyle \begin{align*}&\text{The function }f(x)=524x + 272x^{2} + 84\text{ crosses the x-axis}\\&\text{when it has a value of zero. To find the x values that}\\&\text{satisfy this, we can either factor the equation:}\\&4\cdot (4x + 7)\cdot (17x + 3)\\&\text{or, if this is not intuitive, use the quadratic formula:}\\&\frac{-b\pm\sqrt{b^2-4ac}}{2a}(\text{for equations of the form: }a^2x+bx+c)\\&\frac{-(524)\pm\sqrt{(524)^2-4(272)(84)}}{2(272)}=\frac{-524\pm428}{544}\\&\text{Either way, we find thatthe function crosses the x-axis at}\ x=-\frac{7}{4}\text{ and }-\frac{3}{17}.\end{align*}\)