# SAT Math : Polynomials

## Example Questions

### Example Question #1 : Binomial Denominators

Solve for .

Explanation:

Factor the expression

numerator: find two numbers that add to 2 and multiply to -8 [use 4,-2]

denominator: find two numbers that add to 5 and multiply to -14 [use 7,-2]

new expression:

Cancel the  and cross multiply.

### Example Question #1 : How To Simplify Binomials

If 〖(x+y)〗= 144 and 〖(x-y)〗= 64, what is the value of xy?

16

20

22

18

20

Explanation:

We first expand each binomial to get x2 + 2xy + y2 = 144  and x2 - 2xy + y2 = 64. We then subtract the second equation from the first to find 4xy = 80. Finally, we divide each side by 4 to find xy = 20.

### Example Question #1 : Trinomials

Solve each problem and decide which is the best of the choices given.

What are the zeros of the following trinomial?

Explanation:

First factor out a . Then the factors of the remaining polynomial,

, are  and .

Set everything equal to zero and you get , , and  because you cant forget to set  equal to zero.

### Example Question #21 : Polynomials

Subtract  from .

Explanation:

Step 1: We need to read the question carefully. It says subtract from. When you see the word "from", you read the question right to left.

I am subtracting the left equation from the right equation.

Step 2: We need to write the equation on the right minus the equation of the left.

Step 4: Combine like terms:

Step 5: Put all the terms together, starting with highest degree. The degree of the terms is the exponent. Here, the highest degree is 2 and lowest is zero.

The final equation is

### Example Question #1 : Factoring Polynomials

What is a possible value for x in x2 – 12x + 36 = 0 ?

–6

2

There is not enough information

6

6

Explanation:

You need to factor to find the possible values for x. You need to fill in the blanks with two numbers with a sum of -12 and a product of 36. In both sets of parenthesis, you know you will be subtracting since a negative times a negative is a positive and a negative plus a negative is a negative

(x –__)(x –__).

You should realize that 6 fits into both blanks.

You must now set each set of parenthesis equal to 0.

x – 6 = 0; x – 6 = 0

Solve both equations: x = 6

### Example Question #2 : Factoring Polynomials

If r and t are constants and x2  +rx +6=(x+2)(x+t), what is the value of r?

7

5

6

It cannot be determined from the given information.

5

Explanation:

We first expand the right hand side as x2+2x+tx+2t and factor out the x terms to get x2+(2+t)x+2t. Next we set this equal to the original left hand side to get x2+rx +6=x2+(2+t)x+2t, and then we subtract x2  from each side to get rx +6=(2+t)x+2t. Since the coefficients of the x terms on each side must be equal, and the constant terms on each side must be equal, we find that r=2+t and 6=2t, so t is equal to 3 and r is equal to 5.

### Example Question #3 : Factoring Polynomials

Solve for

Explanation:

First, add 4 to both sides:

Divide both sides by 2:

### Example Question #4 : Factoring Polynomials

Let  and  be integers, such that . If  and , then what is ?

Cannot be determined

Explanation:

We are told that x- y3 = 56. We can factor the left side of the equation using the formula for difference of cubes.

x- y3 = (x - y)(x2 + xy + y2) = 56

Since x - y = 2, we can substitute this value in for the factor x - y.

2(x2 + xy + y2) = 56

Divide both sides by 2.

x2 + xy + y= 28

Because we are trying to find x2 + y2, if we can get rid of xy, then we would have our answer.

We are told that 3xy = 24. If we divide both sides by 3, we see that xy = 8.

We can then substitute this value into the equation x2 + xy + y= 28.

x2 + 8 + y= 28

Subtract both sides by eight.

x2 + y= 20.

ALTERNATE SOLUTION:

We are told that x - y = 2 and 3xy = 24. This is a system of equations.

If we solve the first equation in terms of x, we can then substitute this into the second equation.

x - y = 2

x = y + 2

Now we will substitute this value for x into the second equation.

3(y+2)(y) = 24

Now we can divide both sides by three.

(y+2)(y) = 8

Then we distribute.

y2 + 2y = 8

Subtract 8 from both sides.

y2 + 2y - 8 = 0

We need to factor this by thinking of two numbers that multiply to give -8 but add to give 2. These numbers are 4 and -2.

(y + 4)(y - 2) = 0

This means either y - 4 = 0, or y + 2 = 0

y = -4, or y = 2

Because x = y + 2, if y = -4, then x must be -2. Similarly, if y = 2, then x must be 4.

Let's see which combination of x and y will satisfy the final equation that we haven't used, x- y3 = 56.

If x = -2 and y = -4, then

(-2)3 - (-4)3 = -8 - (-64) = 56. So that means that x= -2 and y = -4 is a valid solution.

If x = 4 and y = 2, then

(4)3 - 2= 64 - 8 = 56. So that means x = 4 and y = 2 is also a valid solution.

The final value we are asked to find is x2 + y2.

If x= -2 and y = -4, then x2 + y= (-2)2 + (-4)2 = 4 + 16 = 20.

If x = 4 and y = 2, then  x2 + y= (4)2 + 22 = 16 + 4 = 20.

Thus, no matter which solution we use for x and y, x2 + y= 20.

### Example Question #5 : Factoring Polynomials

How many negative solutions are there to the equation below?

Explanation:

First, subtract 3 from both sides in order to obtain an equation that equals 0:

The left side can be factored. We need factors of  that add up to and  work:

Set both factors equal to 0 and solve:

To solve the left equation, add 1 to both sides. To solve the right equation, subtract 3 from both sides. This yields two solutions:

Only one of these solutions is negative, so the answer is 1.

### Example Question #6 : Factoring Polynomials

2x + 3y = 5a + 2b        (1)

3x + 2y = 4a – b           (2)

Express x– y2 in terms of a and b

〖–9a〗+ 26ab +〖3b〗2) / 5

(–9a– 27ab +3b2) / 5

(–9a– 28ab –3b2) / 5

–〖9a〗+ 27ab +〖3b〗2) / 5

–〖9a〗+ 26ab –〖3b〗2) / 5