### All SAT Math Resources

## Example Questions

### Example Question #1 : Factoring Polynomials

What is a possible value for x in x^{2} – 12x + 36 = 0 ?

**Possible Answers:**

2

–6

There is not enough information

6

**Correct answer:**

6

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 #1 : Factoring Polynomials

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

**Possible Answers:**

6

5

7

It cannot be determined from the given information.

**Correct answer:**

5

We first expand the right hand side as x^{2}+2x+tx+2t and factor out the x terms to get x^{2}+(2+t)x+2t. Next we set this equal to the original left hand side to get x^{2}+rx +6=x^{2}+(2+t)x+2t, and then we subtract x^{2} 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 #1 : How To Factor A Polynomial

Solve for :

**Possible Answers:**

**Correct answer:**

Solving for simplifies to .

Because there are two 's, there needs to be two solutions, which is why we get .

### Example Question #1 : How To Factor A Polynomial

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

**Possible Answers:**

Cannot be determined

**Correct answer:**

We are told that x^{3 }- y^{3} = 56. We can factor the left side of the equation using the formula for difference of cubes.

x^{3 }- y^{3} = (x - y)(x^{2} + xy + y^{2}) = 56

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

2(x^{2} + xy + y^{2}) = 56

Divide both sides by 2.

x^{2} + xy + y^{2 }= 28

Because we are trying to find x^{2} + y^{2}, 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 x^{2} + xy + y^{2 }= 28.

x^{2} + 8 + y^{2 }= 28

Subtract both sides by eight.

x^{2} + y^{2 }= 20.

The answer is 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

Add y to both sides.

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.

y^{2} + 2y = 8

Subtract 8 from both sides.

y^{2} + 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^{3 }- y^{3} = 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^{3 }= 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 x^{2} + y^{2}.

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

If x = 4 and y = 2, then x^{2} + y^{2 }= (4)^{2} + 2^{2} = 16 + 4 = 20.

Thus, no matter which solution we use for x and y, x^{2} + y^{2 }= 20.

The answer is 20.

### Example Question #1 : Factoring Polynomials

How many negative solutions are there to the equation below?

**Possible Answers:**

**Correct answer:**

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 #2 : Factoring Polynomials

2x + 3y = 5a + 2b (1)

3x + 2y = 4a – b (2)

Express x^{2 }– y^{2} in terms of a and b

**Possible Answers:**

(–9a^{2 }– 28ab –3b^{2}) / 5

–〖9a〗^{2 }+ 26ab –〖3b〗^{2}) / 5

〖–9a〗^{2 }+ 26ab +〖3b〗^{2}) / 5

(–9a^{2 }– 27ab +3b^{2}) / 5

–〖9a〗^{2 }+ 27ab +〖3b〗^{2}) / 5

**Correct answer:**

(–9a^{2 }– 28ab –3b^{2}) / 5

Add the two equations together to yield 5x + 5y = 9a + b, then factor out 5 to get 5(x + y) = 9a + b; divide both sides by 5 to get x + y = (9a + b)/5; subtract the two equations to get x - y = -a - 3b. So, x^{2 }– y^{2} = (x + y)(x – y) = (9a + b)/5 (–a – 3b) = (–[(9a)]^{2 }– 28ab – [(3b)]^{2})/5

### Example Question #1 : Factoring Polynomials

If the polynomial

is divided by

,

what is the remainder?

**Possible Answers:**

**Correct answer:**

By the Remainder Theorem, if a polynomial is divided by a binomial , the remainder is .

Let . Setting , if is divided by , the remainder is , which can be evaluated by setting in the definition of and evaluating:

### Example Question #42 : Variables

Which of the following is a factor of the polynomial ?

**Possible Answers:**

**Correct answer:**

Call

By the Rational Zeroes Theorem, since has only integer coefficients, any rational solution of must be a factor of 54 divided by a factor of 1 - positive or negative. 54 has as its factors 1, 2, 3, 6, 9, 18, 27 , 54; 1 has only itself as a factor. Therefore, the rational solutions of must be chosen from this set:

.

By the Factor Theorem, a polynomial is divisible by if and only if - that is, if is a zero. By the preceding result, we can immediately eliminate and as factors, since 12 and 16 have been eliminated as possible zeroes.

Of the three remaining choices, we can demonstrate that is the factor by evaluating :

, so is a factor.

Of the remaining two choices, and , both can be proved to not be factors by showing that and are both nonzero:

, so is not a factor.

, so is not a factor.

### Example Question #2 : Factoring Polynomials

If the polynomial

is divided by

,

what is the remainder?

**Possible Answers:**

**Correct answer:**

By the Remainder Theorem, if a polynomial is divided by a binomial , the remainder is .

Let . Setting (since ), if is divided by , the remainder is , which can be evaluated by setting in the definition of and evaluating: