### All Algebra II Resources

## Example Questions

### Example Question #3 : Solving Quadratic Functions

Factor the above function to find the roots of the quadratic equation.

**Possible Answers:**

**Correct answer:**

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.

### Example Question #1 : Finding Roots

Find the roots of the function:

**Possible Answers:**

**Correct answer:**

Factor:

Double check by factoring:

Add together:

Therefore:

### Example Question #1 : Finding Zeros Of A Polynomial

Solve for x.

**Possible Answers:**

x = –4, –3

x = 5

x = 5, 2

x = 4, 3

x = –5, –2

**Correct answer:**

x = 5, 2

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

### Example Question #2 : Finding Roots

Solve for x.

**Possible Answers:**

x = 5, 2

x = –5, –2

x = –4

x = 2

x = –4, 4

**Correct answer:**

x = –4

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.

### Example Question #333 : Intermediate Single Variable Algebra

Find the roots of the equation *x*^{2} + 5*x* + 6 = 0

**Possible Answers:**

3 and –3

2 and 3

1 and –3

–5 and 1

–2 and –3

**Correct answer:**

–2 and –3

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

### Example Question #333 : Intermediate Single Variable Algebra

Solve the equation:

**Possible Answers:**

**Correct answer:**

To solve the quadratic equation, , we set the equation equal to zero and then factor the quadratic, . 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 and to obtain the answers and .

### Example Question #3 : How To Simplify Binomials

Solve for :

**Possible Answers:**

**Correct answer:**

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 :

### Example Question #3 : How To Find The Solution To A Binomial Problem

Solve for :

**Possible Answers:**

**Correct answer:**

### Example Question #1 : Finding Roots

Solve for .

**Possible Answers:**

**Correct answer:**

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:

### Example Question #2 : Finding Roots

Solve for :

**Possible Answers:**

**Correct answer:**

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

### All Algebra II Resources

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