# SAT Math : Factoring Polynomials

## Example Questions

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

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

2

6

–6

There is not enough information

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?

5

7

It cannot be determined from the given information.

6

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〗+ 27ab +〖3b〗2) / 5

〖–9a〗+ 26ab +〖3b〗2) / 5

(–9a– 28ab –3b2) / 5

(–9a– 28ab –3b2) / 5

Explanation:

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– y2 = (x + y)(x – y) = (9a + b)/5 (–a – 3b) = (–[(9a)]– 28ab – [(3b)]2)/5

### Example Question #7 : Factoring Polynomials

If the polynomial

is divided by

,

what is the remainder?

Explanation:

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 #8 : Factoring Polynomials

Which of the following is a factor of the polynomial  ?

Explanation:

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 #9 : Factoring Polynomials

If the polynomial

is divided by

,

what is the remainder?