Finding Roots - Algebra 2

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

First factor the equation. Find two numbers that multiply to 24 and sum to -10. These numbers are -6 and -4:

Set both expressions equal to 0 and solve for x:

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!

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

Set both expressions equal to 0 and solve:

and

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!

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

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

1. 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.

1. 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

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

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

1. 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 find the roots, or solutions, of this quadratic equation, first factor the equation.

When factored, it's

.

Then, set each of those expressions equal to 0 and solve for x.

Your solutions are

.

1. Quadratics must be set equal to zero in order to be solved. To do so in this equation, the "8" has to wind up on the left side and combine with any other lone integers. So, multiply out the terms in order to make it possible for the "8" to be added to the other number.

Then combine like terms.

1. Now factor.

1 + 16 = 17

4 + 4 = 8

2 + 8 = 10

1. Pull out common factors, "x" and "8," respectively.

1. Pull out "(x+2)" from both terms.

x = –8, –2

1. Combine like terms and simplify.

No further simplification is possible. The first term has a coefficient that can't be factored away. FOIL requires that all terms be multiplied by each other at some point, so the presence of the coefficient has to be reflected in every step of the factoring.

1. Practically speaking, that means we add an extra step. Multiply the coefficient of the first term by the last term before factoring.

3 * 6 = 18

Factors of 18 include:

1 + 18 = 19

2 + 9 = 11

1. Now pull out the common factor in each of the pairs, "3x" from the first two and "2" from the second two.

1. Pull out the "(x+3)" from both terms.

1. Set both parts equal to zero and solve.

3x + 2 = 0, x = –2/3

x + 3 = 0, x = –3

This is a factoring problem, so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract 128 from both sides to get .

We then notice that all four numbers are divisible by four, so we can simplify the expression to .

Think of the equation in this format to help with the following explanation.

We must then factor to find the solutions for x. To do this we must make a factor tree of c (which is 32 in this case) to find the possible solutions. The possible numbers are 1 * 32, 2 * 16, and 4 * 8.

Since c is negative, we know that our factoring will produce a positive and negative number.

We then look at b to see if the greater number will be positive or negative. Since b is positive, we know that the greater number from our factoring tree will be positive.

We then use addition and subtraction with the factoring tree to find the numbers that add together to equal b. Remember that the greater number is positive and the lesser number is negative in this example.

Positive 8 and negative 4 equal b. We then plug our numbers into the factored form of .

We know that anything multiplied by 0 is equal to 0, so we plug in the numbers for x which make each equation equal to 0. In this case .

.

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 :

Factoring a quadratic equation means doing FOIL backwards. Recall that when you use FOIL, you start with two binomials and end with a trinomial:

Now, we're trying to go the other direction -- starting with a trinomial, and going back to two factors.

Here, -3 is equal to , and -2 is equal to . We can use this information to find out what and are, separately. In other words, we have to find two factors of -3 that add up to -2.

Factors of -3:

• 3*-1 (sum = 2)
• -3*1 (sum = -2)

Thus our factored equation should look like this:

The roots of the quadratic equation are the values of x for which y is 0.

We know that anything times zero is zero. So the entire expression equals zero when at least one of the factors equals zero.

Factor:

Double check by factoring:

Add together:

Therefore:

If we recognize this as an expression with form , with and , we can solve this equation by factoring:

and

and

Looking at this expression, we can see it is of the form , with , , and . Therefore, we can write it in the form :

When asked to solve for , we are really searching for the roots/-intercepts of the equation.

In this particular case, our function is already factored for us leaving us with only a few steps to complete the problem.

Our first step is to set each term equal to , leaving us with...

and

The next step is to use our knowledge of order of operations to simply solve for for each of the above equations...

Subtract from both sides

Divide by

Answer #1, our first root/-intercept

Add to both sides

Answer # 2, our second root/-intercept

This problem requires simplification, order of operations, and knowledge of square roots.

Our goal is to isolate/solve for

Divide by on both sides

Square root both sides

**Remember: DO NOT FORGET THAT WHEN WE TAKE A SQUARE ROOT, WE GET A PLUS/MINUS ANSWER.

or

and

A polynominal with the form multiplies out to , so this leaves two equations:

Factors of 12 are 2, 6, and 3, 4. In this case, because the product is negative, one root is positive and one is negative. Because the sum is negative, the positive root must have a lower absolute value than the negative root. This leaves us with two possibilities: 2, -6 and 3, -4. Plugging into the sum equation shows the roots to be 3, -4. A check into the original polynominal shows

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 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.