To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

So (x + 2)(x + 3) = 0

x = –2 or x = –3

To solve the quadratic equation, \(\displaystyle \small 13= x^{2} -12x\), we set the equation equal to zero and then factor the quadratic, \(\displaystyle \small (x-13)(x+1)=0\). Because these expressions multiply to equal 0, then it must be that at least one of the expressions equals 0. So we set up the corresponding equations \(\displaystyle x-13=0\) and     \(\displaystyle x+1=0\) to obtain the answers \(\displaystyle x=13\) and \(\displaystyle x=-1\). 

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 \(\displaystyle x\):

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

\(\displaystyle (x+3)^2=16\)

\(\displaystyle \sqrt{(x+3)^2}=\sqrt{16}\textup{}\)

\(\displaystyle x+3=\pm4\)

\(\displaystyle x=1\ or\ -7\)

First factor the equation.  Find two numbers that multiply to 24 and sum to -10.  These numbers are -6 and -4: \(\displaystyle (x-6)(x-4)=0\)

Set both expressions equal to 0 and solve for x:

\(\displaystyle (x-6)=0, (x-4)=0\)

\(\displaystyle x=6,4\)

To factor, find two numbers that sum to 5 and multiply to 6.

Check the possible factors of 6:

1 * 6 = 6  

1 + 6 = 7, so these don't work.

2 * 3 = 6  

2 + 3 = 5, so these work!

\(\displaystyle x^{2}+2x+3x+6=0\)

Next, pull out the common factors of the first two terms and then the second two terms:

\(\displaystyle x(x+2)+3(x+2)=0\)

\(\displaystyle (x+2)(x+3)=0\)

Set both expressions equal to 0 and solve:

\(\displaystyle x+2=0\)     

\(\displaystyle x=-2\) 

and

\(\displaystyle x+3=0\)

\(\displaystyle x=-3\)

Add 8 to both sides to set the equation equal to 0:

\(\displaystyle \small 3h^2+10h+8=0\)

To factor, find two integers that multiply to 24 and add to 10.  4 and 6 satisfy both conditions. Thus, we can rewrite the quadratic of three terms as a quadratic of four terms, using the the two integers we just found to split the middle coefficient:

\(\displaystyle \small 3h^2+6h+4h+8=0\)

Then factor by grouping:

\(\displaystyle \small 3h(h+2)+4(h+2)=0 \rightarrow (h+2)(3h+4)=0\)

Set each factor equal to 0 and solve:

\(\displaystyle h=-2\)

and

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

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}.

First we remember that "find the roots" means "find the values of x for which this expression equals 0." So we set the expression = 0 and approach solving as normal.

\(\displaystyle 3x^2 -4x - 15 = 0\)

Since solving this by sight is difficult, we'll use composition, multiplying a by c and finding factors which add to b.

\(\displaystyle 3 * -15 = -45\)

\(\displaystyle -45 = -9 * 5\)

\(\displaystyle -9 + 5 = -4\)

So -9 and 5 will work; we will use them to rewrite -4x as -9x + 5x and then factor by grouping.

\(\displaystyle 3x^2 -9x + 5x - 15 = 0\)

\(\displaystyle (3x^2 -9x) + (5x - 15) = 0\)

We identify common factors to "pull" out of each group.

\(\displaystyle ((3x)x -3(3x)) + ((5)x - (5)3) = 0\)

\(\displaystyle 3x(x -3) + 5(x - 3) = 0\)

And now we factor out x-3.

\(\displaystyle (x-3)(3x+5) = 0\)

Setting each factor equal to 0 lets us solve for x. 

\(\displaystyle x - 3 = 0\)

\(\displaystyle x = 3\)

\(\displaystyle 3x + 5 = 0\)

\(\displaystyle 3x = -5\)

\(\displaystyle x = -\frac{5}{3}\)

So our solutions are x = -5/3 and x = 3, which we write as x = {-5/3, 3}.

Notice in this question there are only two terms, the exponent value and the constant value. There is also the negative sign between the two. When we look at each number we see that each are a perfect square. Due to the negative sign between the two, this type of quadratic expression can also be written as a difference of squares. We look at the exponential term and see it is

 \(\displaystyle 16x^2\)

The perfect square factors of this term are \(\displaystyle 4x\)  and \(\displaystyle 4x\) .

Now we look at the constant term

\(\displaystyle 144\)

The perfect square factors of this term are \(\displaystyle 12\)  and \(\displaystyle 12\)

Now to combine these into the binomial factor form we need to remember it is the difference of perfect squares meaning we will have one subtraction sign and one adding sign, so we get the following:

\(\displaystyle (4x-12)(4x+12)\)

From here we solve each binomial for x. To do this we set each binomal to zero and solve for x.

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

\(\displaystyle 4x=12\)                        \(\displaystyle 4x=-12\)

\(\displaystyle x=\frac{12}{4}\)                          \(\displaystyle x=\frac{-12}{4}\)

\(\displaystyle x=3\)                             \(\displaystyle x=-3\)