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# Using Factoring Practice Questions

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Updated on Sep 23, 2011

## Introduction

This set of practice questions will present polynomial expressions for you to factor. In this lesson, the coefficients of the second-degree terms will often be whole numbers greater than 1. This will complicate the process of factoring by adding more possibilities to check. In some cases, you will find that you can factor using more than one of the three methods of factoring polynomials on the given expression.

Always look to factor algebraic expressions using the greatest common factor method first. Then analyze the remaining expression to determine if other factoring methods can be used. The three methods for factoring polynomial expressions are:

1. Greatest common factor method
2. Difference of two perfect squares method
3. Trinomial method

When presented with a polynomial with a coefficient greater than 1 for the second-degree term, use the trinomial factor form (ax ± ( ))(bx ± ( )) where a · b = the coefficient of the second-degree term. List the factors of the numerical term of the trinomial and consider the choices of factors and signs that will result in the correct trinomial factorization.

After choosing terms to try in the trinomial factors form, use FOIL to check your guesses for the trinomial factors. You will want to do a partial check by first completing the O and the I part of FOIL to determine if you have the first-degree term right.

## Tips for Using Factoring

When factoring a trinomial expression, first determine the signs that will be used in the two factors.

Next, list the possible factors of the second-degree term.

Then list the factors of the numerical term.

Finally, place the factors into the trinomial factor form in all possible ways and use FOIL to check for the correct factorization.

Be systematic in your attempts to be sure you try all possible choices. You will become better at factoring as you learn to look for the combinations of factors that will give you the required results for the first-degree term.

## Practice Questions

Factor the following expressions.

1. 2x2 + 7x + 6
2. 3x2 + 13x + 12
3. 5x2 – 14x – 3
4. 9x2 + 15x + 4
5. 9x2 + 34x – 8
6. 3x2 – 3x – 18
7. 4a2 – 16a – 9
8. 6a2 – 13a – 15
9. 6a2 – 5a – 6
10. 16y2 – 100
11. 6x2 + 15x – 36
12. 4bc2 + 22bc – 42b
13. 2a6 + a3 – 21
14. 6a2x – 39ax – 72x
15. 8x2 – 6x – 9
16. 5c2 – 9c – 2
17. 9x3 – 4x
18. 8r2 + 46r + 63
19. 4x4 – 37x2 + 9
20. 12d2 + 7d – 12
21. 4xy3 + 6xy2 – 10xy
22. 4ax2 – 38ax – 66a
23. 3c2 + 19c – 40
24. 2a2 + 17a – 84
25. 4x4 + 2x2 – 30

Numerical expressions in parentheses like this [ ] are operations performed on only part of the original expression. The operations performed within these symbols are intended to show how to evaluate the various terms that make up the entire expression.

Expressions with parentheses that look like this ( ) contain either numerical substitutions or expressions that are part of a numerical expression. Once a single number appears within these parentheses, the parentheses are no longer needed and need not be used the next time the entire expression is written.

When two pair of parentheses appear side by side like this ( )( ), it means that the expressions within are to be multiplied.

Sometimes parentheses appear within other parentheses in numerical or algebraic expressions. Regardless of what symbol is used, ( ), { }, or [ ], perform operations in the innermost parentheses first and work outward.

Underlined expressions show the original algebraic expression as an equation with the expression equal to its simplified result.

1. Both signs in the trinomial are positive, so use positive signs in the trinomial factor form.
(ax + ( ))(bx + ( ))
2. The factors of the second-degree term are 2x2 = (2x)(x).

The numerical term 6 = (1)(6) = (2)(3).

You want to get 7x from adding the result of the Outer and Inner multiplications when using FOIL. You could make the following guesses for the factors of the original expression.

(2x + (1))(x + (6))
(2x + (6))(x + (1))
(2x + (2))(x + (3))
(2x + (3))(x + (2))

Now just consider the results of the Outer and Inner products of the terms for each guess. The one that results in a first-degree term of 7x is the factorization you want to fully check.

(2x + (1))(x + (6)) will result in Outer product plus Inner product:

2x(6) + (1)x = 12x + x = 13x.

(2x + (6))(x + (1)) will result in Outer product plus Inner product:

2x(1) + (6)x = 2x + 6x = 8x.

(2x + (2))(x + (3)) will result in Outer product plus Inner product:

2x(3) + (2)x = 6x + 2x = 8x.

(2x + (3))(x + (2)) will result in Outer product plus Inner product:

2x(2) + (3)x = 4x + 3x = 7x.

Place the factors in the trinomial factor form so that the product of the outer terms (2x)(2) = 4x and the product of the inner terms (3)(x) = 3x. That way, 4x + 3x = 7x, the middle term of the trinomial.

(2x + (3))(x + (2))
 Check using FOIL. First— –(2x)(x) = 2x2 Outer— –(2x)(2) = 4x Inner— –(3)(x) = 3x Last— –(3)(2) = 6 Add the products of multiplication using FOIL. 2x2 + 4x + 3x + 6 = 2x2 + 7x + 6 The factors check out.
• Both signs in the trinomial are positive, so use positive signs in the trinomial factor form.
(ax + ( ))(bx + ( ))

The factors of the second-degree term are 3x2 = (3x)(x).

The factors of the numerical term 12 = (1)(12) = (2)(6) = (3)(4).

You want to get 13x from adding the result of the Outer and Inner multiplications when using FOIL. Place the factors in the trinomial factor form so that the product of the outer terms (3x)(3) = 9x and the product of the inner terms (4)(x) = 4x. Then 9x + 4x = 13x, the middle term of the trinomial.

(3x + 4)(x + 3)
 Check using FOIL. First#8212; –(3x)(x) = 3x2 Outer#8212; –(3x)(3) = 9x Inner#8212; –(4)(x) = 4x Last#8212; –(4)(3) = 12 Add the products of multiplication using FOIL. 3x2 + 9x + 4x + 12 = 3x2 + 13x + 12 The factors check out.
• Both signs in the trinomial are negative. To get a negative sign for the numerical term, the signs in the factors must be + and –.
(ax + ( ))(bx – ( ))

The factors of the second-degree term are 5x2 = (5x)(x).

The factors of the numerical term 3 = (1)(3).

When you multiply the Outer and Inner terms of the trinomial factors, the results must add up to be –14x. Multiplying, 5x(3) = 15x, and 1x(+1) = +1x. Adding (15x) + (+1x) = 14x. Place those terms into the trinomial factor form.

(5x + (1))(x – (3))
 Check using FOIL. First#8212; –(5x)(x) = 5x2 Outer#8212; –(5x)(–3) = –15x Inner#8212; –(1)(x) = x Last#8212; –(1)(–3) = –3 Add the products of multiplication using FOIL. 5x2 – 15x + 1x – 3 = 5x2 – 14x – 3 The factors check out.
• Both signs in the trinomial are positive, so use positive signs in the trinomial factor form. (ax + ( ))(bx + ( ))

The factors of the second-degree term 9x2 = (9x)(x) or 9x2 = (3x)(3x).

The factors of the numerical term 4 = (1)(4) or 4 = (2)(2)

To get 15x from adding the result of the Outer and Inner multiplications when using FOIL, place the factors in the trinomial factor form so that the product of the outer terms (3x)(1) = 3x and the product of the inner terms (4)(3x) = 12x. Then 3x + 12x = 15x, the middle term of the trinomial.

 (3x + 4)(3x + 1) Check using FOIL. First#8212; –(3x)(3x) = 9x2 Outer#8212; –(3x)(1) = 3x Inner#8212; –(4)(3x) = 12x Last#8212; –(4)(1) = 4 The factors check out.
• The sign of the numerical term is negative. So the signs in the trinomial factor form will have to be + and –. That is the only way to get a negative sign by multiplying the Last terms when checking with FOIL. (ax + ( ))(bx – ( ))

The factors of the second-degree term are 9x2 = (9x)(x) or 9x2 = (3x)(3x).

The factors of the numerical term 8 = (1)(8) or (2)(4). Let's try putting in factors in the trinomial factor form and see what we get. (9x + 1)(x – 8)

Using FOIL to check, we get 9x2 – 72x + 1x – 8 = 9x2 – 71x – 8.

No, that doesn't work. You are looking for a positive x term in the middle of the expression. Changing position of the signs would help but not with these factors because the term would be +71x. Try different factors of 8.

 (9x – 2)(x + 4) Check using FOIL.

There it is! And on only the second try. Be persistent and learn from your mistakes.

• The three terms have a common factor of 3. You can factor out 3 and represent the trinomial as 3(x2x – 6). Now factor the trinomial in the parentheses, and don't forget to include the factor 3 when you are done.
 The sign of the numerical term is negative. So the signs in the trinomial factor form will have to be + and –, because that is the only way to get a negative sign when multiplying the Last terms when checking with FOIL. (ax + ( ))(bx – ( ))

The factors of the second-degree term are 3x2 = (3x)(x).

 The sign of the second term is negative. That tells you that the result of adding the products of the Outer and Inner terms of the trinomial factors must result in a negative sum for the x term. The factors of the numerical term 6 are (1)(6) or (2)(3). Put the + with the 2 and the – with the 3. (x + 2)(x – 3) Check using FOIL. First#8212; –(x)(x) = x2 Outer#8212; –(x)(–3) = –3x Inner#8212; –(2)(x) = 2x Last#8212; –(2)(–3) = –6 The factors check out. (x + 2)(x – 3) = x2 – 3x + 2x – 6 = x2 – x – 6

Include the common factor of 3 so that .

• Both signs in the trinomial expression are negative. To get a negative sign for the numerical term, the signs within the trinomial factors must be + and –. (ax + ( ))(bx – ( ))

The factors of the second-degree term 4a2 = (4a)(a) or 4a2 = (2a)(2a).

The factors of the numerical term 9 are (1)(9) or (3)(3).

 The coefficient of the first-degree term is 2 less than 18. You can multiply 2a and (9) to get 18a leaving the factors 2a and (1) to get a 2a. Use this information to place factors within the trinomial factor form. (2a + 1)(2a – 9) Check using FOIL. First#8212; –(2a)(2a) = 4a2 Outer#8212; –(2a)(–9) = –18a Inner#8212; –(1)(2a) = 2a Last#8212; –(1)(–9) = –9

The result of multiplying the factors is .

• Both signs in the trinomial expression are negative. To get a negative sign for the numerical term, the signs within the trinomial factors must be + and –. (ax + ( ))(bx – ( ))

The factors of the second-degree term 6a2 = (6a)(a) or (2a)(3a).

The factors of the numerical term 15 are (1)(15) or (3)(5).

We can predict that 13 = 18 – 5. The factors (6a)(3) = 18a. The remaining factors (a)(5) = 5a. But we need the 13 to be negative, so arrange the 6a and the (3) so their product is –18a.

(6a + 5)(a – 3)

Check using FOIL.

First—(6a)(a) = 6a2

Outer—(6a)(–3) = –18a

Inner—(5)(a) = 5a

Last—(5)(–3) = –15

Combining the results of multiplying using FOIL results in

.

The factors check out.

• Both signs in the trinomial expression are negative. To get a negative sign for the numerical term, the signs within the trinomial factors must be + and –. (ax + ( ))(bx – ( ))

The factors of the second-degree term 6a2 = (6a)(a) or 6a2 = (2a)(3a).

The factors of the numerical term 6 are (6)(1) or (2)(3).

The trinomial looks balanced with a 6 on each end and a 5 in the middle.

 Try a balanced factor arrangement. (3a + 2)(2a – 3) Check using FOIL. First—–(3a)(2a) = 6a2 Outer—–(3a)(–3) = –9a Inner—–(2)(a) = 2a Last—–(2)(–3) = –6

Combining the results of multiplying using FOIL results in

Didn't that work out nicely? A sense of balance can be useful.

• This expression is the difference between two perfect squares. Using the form for the difference of two perfect squares gives you the factors (4y + 10)(4y – 10).

However, there is a greatest common factor that could be factored out first to leave 4(4y2 – 25). Now you need only factor the difference of two simpler perfect squares.

4(4y2 – 25) = 4(2y + 5)(2y – 5)

The first factorization is equivalent to the second because you can factor out two from each of the factors. This will result in (2)(2)(2y + 5)(2y – 5) or 4(2y + 5)(2y – 5).. When factoring polynomials, watch for the greatest common factors first.

• The terms of the trinomial have a greatest common factor of 3. So the term 3(2x2 + 5x – 12) will simplify the trinomial factoring. You need only factor the trinomial within the parentheses.

The sign of the numerical term is negative. So the signs in the trinomial factor form will have to be + and –. (ax + ( ))(bx – ( ))

The factors of the second-degree term are 2x2 = (2x)(x).

The factors of the numerical term 12 are (1)(12) or (2)(6) or (3)(4).

The factors (2x)(4) = 8x and the remaining factors (x)(3) = 3x. It's clear that 8x – 3x = 5x. Use those factors in the trinomial factor form.

(2x – 3)(x + 4)

Check using FOIL.

First—(2x)(x) = 2x2

Outer—(2x)(4) = 8x

Inner—(–3)(x) = –3x

Last—(–3)(4) = –12

The result of multiplying factors is 2x2 + 8x – 3x – 12 = 2x2 + 5x – 12.

Now include the greatest common factor of 3 for the final solution.

• Each term in the polynomial has a common factor of 2b. The resulting expression looks like this: 2b(2c2 + 11c – 21)

The sign of the numerical term is negative. So the signs in the trinomial factor form will have to be + and – because that is the only way to get a negative sign when multiplying the Last terms when checking with FOIL. (ax + ( ))(bx – ( ))

The factors of the second-degree term 2c2 = (2c)(c).

The factors of the numerical term 21 are (1)(21) or (3)(7). The factors (2c)(7) = 14c and the associated factors (c)(3) = 3c. Place these factors in the trinomial factor form so that the result of the Outer and Inner products when using FOIL to multiply are +14c and –3c.       (2c – 3)(c + 7)

Check using FOIL.

First—(2c)(c) = 2c2

Outer—(2c)(7) = 14c

Inner—(–3)(c) = –3c

Last—(–3)(7) = –21

The product of the factors is

(2c – 3)(c + 7) = 2c2 + 14c – 3c – 21 = 2c2 + 11c – 21.

Now include the greatest common factor term.

• This expression appears to be in the familiar trinomial form, but what's with those exponents? Think of a6 = (a3)2. Then the expression becomes 2(a3)2 + (a3) – 21. Now you factor like it was a trinomial expression. The sign of the numerical term is negative. So the signs in the trinomial factor form will have to be + and –. (ax + ( ))(bx – ( ))

The second-degree term 2(a3)2 = 2(a3)(a3).

The factors of the numerical term 21 are (1)(21) or (3)(7). The factors (a3)(7) = 7(a3) and the factors (2(a3))(3) = 6(a3). The difference of 7 and 6 is 1. Place these factors in the trinomial factor form so that the first degree term is 1(a3).

(a3 – 3)(2a3 + 7)

Check using FOIL.

First—(a3)(2a3) = 2a6

Outer—(a3)(7) = 7a3

Inner—(–3)(2a3) = –6a3

Last—(–3)(7) = –21

The product of the factors is

The factors of the trinomial are correct.

• The greatest common factor of the terms in the trinomial expression is 3x. Factoring 3x out results in the expression 3x(2a2 – 13a – 24). Factor the trinomial expression inside the parentheses.

The sign of the numerical term is negative. So the signs in the trinomial factor form will have to be + and –. (ax + ( ))(bx – ( ))

The factors of the term 2a2 = (2a)(a).

The factors of the numerical term 24 are (1)(24) or (2)(12) or (3)(8) or (4)(6).

The factors (2a)(8) = 16a, and the related factors (a)(3) = 3a. The difference of 16 and 3 is 13. Place these numbers in the trinomial factor form, and check the expression using FOIL.

(2a + 3)(a – 8)

First—(2a)(a) = 2a2

Outer—(2a)(–8) = –16a

Inner—(3)(a) = 3a

Last—(3)(–8) = –24

The result is (2a + 3)(a – 8) = 2a2 – 16a + 3a – 24 = 2a2 – 13a – 24.

Now include the greatest common factor if 3x.

• The numerical term of the trinomial has a negative sign, so the signs within the factors of the trinomial will be a + and –. (ax + ( ))(bx – ( ))

The factors of the second-degree term are 8x2 = (x)(8x) or 8x2 = (2x)(4x).

The numerical term of the trinomial 9 has factors of (1)(9) or (3)(3).

What combination will result in a –6x when the Outer and Inner products of the multiplication of the trinomial factors are added together? Consider just the coefficients of x and the numerical term factors. The numbers 2(3) = 6, and the corresponding 4(3) = 12. The difference between 12 and 6 is 6. So use the second-degree term factors (2x)(4x) and the numerical factors (3)(3).

(2x – 3)(4x + 3)

Check using FOIL.

First—(2x)(4x) = 8x2

Outer—(2x)(3) = 6x

Inner—(–3)(4x) = –12x

Last—(–3)(3) = –9

The product of the factors

• The numerical term of the trinomial has a negative sign, so the signs within the factors of the trinomial will be a + and –. (ax + ( ))(bx – ( ))

The only factors of the second-degree term are (c)(5c).

The numerical term of the trinomial 2 has factors of (1)(2). What combination will result in a –9c when the Outer and Inner products of the trinomial factors are added together?

Our choices are (5)(1)c + (1)(–2)c, (5)(–1)c + (1)(2), (5)(2)c + (1)(–1)c, (5)(–2)c + (1)(1)c. The last of these is equal to the desired –9c, which gives the factoring (5c + 1)(c – 2). Check using FOIL to multiply terms. (5c + 1)(c – 2)

Check using FOIL.

First—(5c)(c) = 5c2

Outer—(5c)(–2) = –10c

Inner—(1)(c) = c

Last—(1)(–2) = –2

The product of the factors .

• The terms of the expression have a greatest common factor of x. Factoring x out of the expression results in x(9x2 – 4).

The expression inside the parentheses is the difference of two perfect squares. Factor that expression using the form for the difference of two perfect squares. Include the greatest common factor to complete the factorization of the original expression.

9x2 = (3x)2
4 = 22

Using the form, the factorization of the difference of two perfect squares is (3x – 2)(3x + 2).

Check using FOIL.

First—(3x)(3x) = 9x2

Outer—(3x)(2) = 6x

Inner—(–2)(3x) = –6x

Last—(–2)(2) = –4

Include the greatest common factor x in the complete factorization.

• The terms in the trinomial expression are all positive, so the signs in the trinomial factor form will be positive. (ax + ( ))(bx + ( ))

The factors of the second-degree term are 8r2 = (r)(8r) or (2r)(4r).

The numerical term 63 has the factors (1)(63) or (3)(21) or (7)(9).

You need two sets of factors that when multiplied and added will result in a 46. Let's look at the possibilities using the 2r and 4r. 2r(1) + 4r(63) = 254r is too much. 2r(3) + 4r(21) = 87r is still too much. Try 2r(21) and 4r(3) = 54r. Getting closer. 2r(7) + 4r(9) = 50r. Nope. Now try 2r(9) + 4r(7) = 46r. Bingo!

(2r + 7)(4r + 9)

Check using FOIL.

First—(2r)(4r) = 8r2

Outer—(2r)(9) = 18r

Inner—(4r)(7) = 28r

Last—(7)(9) = 63

Add the result of the multiplication. 8r2 + 18r + 28r + 63 = 8r2 + 46r + 63

The factors check out.

• When you think of x4 = (x2)2, you can see that the expression is a trinomial that is easy to factor.

The numerical term is positive, so the signs in the trinomial factor form will be the same. The sign of the first-degree term is negative, so you will use two – signs. (ax – ( ))(bx – ( ))

The factors of the second-degree term are 4x4 = (x2)(4x2) or (2x2)(2x2).

The numerical term 9 has (1)(9) or (3)(3) as factors.

What combination will result in a total of 37 when the Outer and Inner products are determined? 4x2(9) = 36x2, 1x2 (1) = 1x2 and 36x2 + 1x2 = 37x2. Use these factors in the trinomial factor form.

(4x2 – 1)(x2 – 9)

Check using FOIL and you will find

(4x2 – 1)(x2 – 9) = 4x4 – 36x2x2 + 9 = 4x4 – 37x2 + 9.

Now you need to notice that the factors of the original trinomial expression are both factorable. Why? Because they are both the difference of two perfect squares.

Use the factor form for the difference of two perfect squares for each factor of the trinomial.

(4x2 – 1) = (2x + 1)(2x – 1)
(x2 – 9) = (x + 3)(x – 3)

Put the factors together to complete the factorization of the original expression.

• The negative sign in front of the numerical term tells you that the signs of the trinomial factors will be + and –. (ax + ( ))(bx – ( ))

This expression has a nice balance to it with 12 at the extremities and a modest 7 in the middle. Let's guess at some middle of the road factors to plug in. Use FOIL to check.

(4d + 3)(3d – 4)

Using FOIL, you find

(4d + 3)(3d – 4) = 12d2 – 16d + 9d – 12 = 12d2 – 7d – 12.

Those are the right terms but the wrong signs. Try changing the signs around.

(4d – 3)(3d + 4)

Multiply the factors using FOIL.

This is the correct factorization of the original expression.

• Each term in the expression has a common factor of 2xy. When factored out, the expression becomes 2xy(2y2 + 3y – 5). Now factor the trinomial in the parentheses.

The last sign is negative, so the signs within the factor form will be a + and –.

(ax + ( ))(bx – ( ))

The factors of the second-degree term are 2y2 = y(2y).

The numerical term 5 has factors (5)(1).

Place the factors of the second degree and the numerical terms so that the result of the Outer and Inner multiplication of terms within the factor form of a trinomial expression results in a +3x.

(2y + 5)( y – 1)

Multiply using FOIL. (2y + 5)(y – 1) = 2y2 – 2y + 5x – 5 = 2y2 + 3x – 5

The factors of the trinomial expression are correct. Now include the greatest common factor to complete the factorization of the original expression.

• The terms of the trinomial have a greatest common factor of 2a. When factored out, the resulting expression is 2a(2x2 – 19x – 33). The expression within the parentheses is a trinomial and can be factored.

The signs within the terms of the factor form will be + and – because the numerical term has a negative sign. Only a (+)(–) = (–). (ax + ( ))(bx – ( ))

The factors of the second-degree term are 2a2 = a(2a).

The numerical term 33 has (1)(33) or (3)(11) as factors.

Since 2a(11) = 22a, and a(3) = 3a, and 22a – 3a = 19a, use those factors in the trinomial factor form so that the result of the multiplication of the Outer and Inner terms results in –19x.

(2x + 3)(x – 11)

Check using FOIL. (2x + 3)(x – 11) = 2x2 – 22x + 3x – 33 = 2x2 – 19x – 33

The factorization of the trinomial factor is correct. Now include the greatest common factor of the original expression to get the complete factorization of the original expression.

• The signs within the terms of the factor form will be + and – because the numerical term has a negative sign. (ax + ( ))(bx – ( ))

The factors of the second-degree term are 3c2 = c(3c).

The numerical term 40 has (1)(40) or (2)(20) or (4)(10) or (5)(8) as factors.

You want the result of multiplying and then adding the Outer and Inner terms of the trinomial factor form to result in a +19c when the like terms are combined. Using trial and error, you can determine that 3c(8) = 24c, and c(5) = 5c, and 24c – 5c = 19c. Use those factors in the factor form in such a way that you get the result you seek.

(3c – 5)(c + 8) = 3c2 + 24c – 5c – 40 = 3c2 + 19c – 40

The complete factorization of the original expression is (3c – 5)(c + 8).

• The signs within the terms of the factor form will be + and – because the numerical term has a negative sign. (ax + ( ))(bx – ( ))

The factors of the second-degree term are 2a2 = a(2a).

The numerical term 84 has (1)(84) or (2)(42) or (3)(28) or (4)(21) or (6)(14) or (7)(12) as factors. You want the result of multiplying and then adding the Outer and Inner terms of the trinomial factor form to result in a +17a when the like terms are combined.

2a(12) = 24a, and a(7) = 7a, and 24a – 7a = 17a. Use the factors 2a and a as the first terms in the factor form and use (12) and (7) as the numerical terms. Place them in position so you get the result that you want.

(2a – 7)(a + 12) = 2a2 + 24a – 7a – 84 = 2a2 + 17a – 84

The complete factorization of the original expression is (2a – 7)(a + 12).

• This expression is in trinomial form. If you think of the variable as x2, you can see that the expression is in the trinomial form. Use x2 where you usually put a first-degree variable. The trinomial you will be factoring looks like this: 4(x2)2 + 2(x2) – 30.

The signs within the terms of the factor form will be + and – because the numerical term has a negative sign.

(ax + ( ))(bx – ( ))

The term (4x2)2 can be factored as (4x2)2 = x2(4x2) or (2x2)(2x2).

The numerical term 30 can be factored as (1)(30) or (2)(15) or (3)(10) or (5)(6).

The factors (4x2)(3) = 12x2 and x2(10) = 10x2 will give you 12x2 – 10x2 = 2x2 when you perform the Inner and Outer multiplications and combine like terms using FOIL with the terms in the trinomial factor form. The factors of the expression will be (4x2 – 10)(x2 + 3).

Check using FOIL.

(4x2 – 10)(x2 + 3) = 4x4 + 12x2 – 10x2 – 30 = 4x4 + 2x2 – 30

The expression (4x2 – 10)(x2 + 3) is the correct factorization of the original expression. However, the first factor has a greatest common factor of 2. So a complete factorization would be 2(2x2 – 5)(x2 + 3).

Did you notice that you could have used the greatest common factor method to factor out a 2 from each term in the original polynomial? If you did, you would have had to factor the trinomial expression 2x4 + x2 – 15 and multiply the result by the factor 2 to equal the original expression. Let's see:

It all comes out the same, but if you left the factor of 2 in the term (4x2 – 10), you wouldn't have done a complete factorization of the original trinomial expression.

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