### All PSAT Math Resources

## Example Questions

### Example Question #1 : Variables

Factor the following variable

(x^{2} + 18x + 72)

**Possible Answers:**

(x + 18) (x + 72)

(x + 6) (x + 12)

(x – 6) (x + 12)

(x + 6) (x – 12)

(x – 6) (x – 12)

**Correct answer:**

(x + 6) (x + 12)

You need to find two numbers that multiply to give 72 and add up to give 18

easiest way: write the multiples of 72:

1, 72

2, 36

3, 24

4, 18

6, 12: these add up to 18

(x + 6)(x + 12)

### Example Question #1 : Factoring

Factor 9*x*^{2} + 12*x* + 4.

**Possible Answers:**

(3*x* + 2)(3*x* – 2)

(3*x* – 2)(3*x* – 2)

(9*x* + 4)(9*x* + 4)

(9*x* + 4)(9*x* – 4)

(3*x* + 2)(3*x* + 2)

**Correct answer:**

(3*x* + 2)(3*x* + 2)

Nothing common cancels at the beginning. To factor this, we need to find two numbers that multiply to 9 * 4 = 36 and sum to 12. 6 and 6 work.

So 9*x*^{2} + 12*x* + 4 = 9*x*^{2} + 6*x* + 6*x* + 4

Let's look at the first two terms and last two terms separately to begin with. 9*x*^{2} + 6*x* can be simplified to 3*x*(3*x* + 2) and 6*x* + 4 can be simplified into 2(3*x* + 2). Putting these together gets us

9*x*^{2} + 12*x* + 4

= 9*x*^{2} + 6*x* + 6*x* + 4

= 3*x*(3*x* + 2) + 2(3*x* + 2)

= (3*x* + 2)(3*x* + 2)

This is as far as we can factor.

### Example Question #2 : Factoring Polynomials

If , and , what is the value of ?

**Possible Answers:**

–8

6

8

0

–6

**Correct answer:**

8

The numerator on the left can be factored so the expression becomes , which can be simplified to

Then you can solve for by adding 3 to both sides of the equation, so

### Example Question #2 : Factoring

Solve for *x*:

**Possible Answers:**

**Correct answer:**

First, factor.

Set each factor equal to 0

Therefore,

### Example Question #1 : Factoring

When is factored, it can be written in the form , where , , , , , and are all integer constants, and .

What is the value of ?

**Possible Answers:**

**Correct answer:**

Let's try to factor x^{2} – y^{2} – z^{2} + 2yz.

Notice that the last three terms are very close to y^{2} + z^{2} – 2yz, which, if we rearranged them, would become y^{2} – 2yz+ z^{2}. We could factor y^{2} – 2yz+ z^{2 } as (y – z)^{2}, using the general rule that p^{2} – 2pq + q^{2 } = (p – q)^{2} .

So we want to rearrange the last three terms. Let's group them together first.

x^{2} + (–y^{2} – z^{2} + 2yz)

If we were to factor out a –1 from the last three terms, we would have the following:

x^{2} – (y^{2} + z^{2} – 2yz)

Now we can replace y^{2} + z^{2} – 2yz with (y – z)^{2}.

x^{2} – (y – z)^{2}

This expression is actually a differences of squares. In general, we can factor p^{2} – q^{2} as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.

x^{2} – (y – z)^{2 }= (x – (y – z))(x + (y – z))

Now, let's distribute the negative one in the trinomial x – (y – z)

(x – (y – z))(x + (y – z))

(x – y + z)(x + y – z)

The problem said that factoring x^{2} – y^{2} – z^{2} + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.

(x – y + z)(x + y – z) fits this form. This means that a = 1, b = –1, c = 1, d = 1, e = 1, and f = –1. The sum of all of these is 2.

The answer is 2.

### Example Question #1 : Factoring

Factor and simplify:

**Possible Answers:**

**Correct answer:**

is a difference of squares.

The difference of squares formula is .

Therefore, = .

### Example Question #3 : Factoring

Factor:

**Possible Answers:**

**Correct answer:**

We can first factor out :

This factors further because there is a difference of squares:

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