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!

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

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

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

1) First step of solving any equation: combine like terms. With quadratics, the easiest step to take is to set the expression equal to zero.

2) There are two ways to do this problem. The first and most intuitive method is standard factoring.

16 + 1 = 17

8 + 2 = 10

4 + 4 = 8

3) Then follow the usual steps, pulling out the common factor from both pairs, "x" from the first and "4" from the second.

4) Pull out the "(x+4)" to wind up with:

5) Set each term equal to zero.

x + 4 = 0, x = –4

But there's a shortcut! Assuming the terms are arranged by descending degree (i.e., ), and the third term is both a perfect square whose square root is equal to half of the middle term, mathematicians use a little trick. In this case, the square root of 16 is 4. 4 * 2=8, so the trick will work. Take the square root of the first and last term, then stick a plus sign in between them and square the parentheses. 

And x, once again, is equal to –4.

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

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

Finally, you set each binomial equal to 0 and solve for :

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

Then factor by grouping:

Set each factor equal to 0 and solve:

and

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.

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.

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

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.

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

Setting each factor = 0 we can find the solutions.

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.

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

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

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

And now we factor out x-3.

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

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

The perfect square factors of this term are   and  .

Now we look at the constant term

The perfect square factors of this term are   and 

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:

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

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

The perfect square factors of this term are  and  .

Now we look at the constant term

The perfect square factors of this term are   and  

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:

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

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

The perfect square factors of this term are  and .

Now we look at the constant term

The perfect square factors of this term are  and . 

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:

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

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 term we see that each is 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

The perfect square factors of this term are  and .

Now we look at the constant term

The perfecct square factors of this term are  and .

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:

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