### All PSAT Math Resources

## Example Questions

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

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

**Possible Answers:**

2

6

–6

There is not enough information

**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:**

It cannot be determined from the given information.

6

5

7

**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 : Factoring Polynomials

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 #4 : 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 #13 : New Sat Math Calculator

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 #1 : How To Factor A Polynomial

How many of the following are *prime* factors of the polynomial ?

(A)

(B)

(C)

(D)

**Possible Answers:**

None

Three

Four

One

Two

**Correct answer:**

One

can be seen to fit the pattern

:

where

can be factored as , so

, as the sum of squares, is a prime polynomial, so the complete factorization is

,

making the only prime factor, and "one" the correct choice.

### Example Question #11 : Variables

Completely factor the following expression:

**Possible Answers:**

**Correct answer:**

To begin, factor out any like terms from the expression. In this case, the term can be pulled out:

Next, recall the difference of squares:

Here, and .

Thus, our answer is

.

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

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 }+ 26ab –〖3b〗^{2}) / 5

(–9a^{2 }– 28ab –3b^{2}) / 5

(–9a^{2 }– 27ab +3b^{2}) / 5

–〖9a〗^{2 }+ 27ab +〖3b〗^{2}) / 5

〖–9a〗^{2 }+ 26ab +〖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

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