Factoring by Grouping

Question 1

Factor 10b^3 + 15b^2 + 14b + 21 completely.

Accepted Answer

(2b + 3)(5b^2 + 7)

Answer

5b^2(2b + 3) + 7(2b + 3)

Answer

(5b + 7)(2b^2 + 3)

Answer

(2b + 3)(5b^2 - 7)

Question 2

Does 5t^3 + 35t^2 + 9t + 63 factor by grouping?

Accepted Answer

yes, both pairs leave (t + 7), so it factors as (t + 7)(5t^2 + 9)

Answer

no, the two pairs leave different binomials, so this one does not group at all

Answer

yes, but only as far as 5t^2(t + 7) + 9(t + 7), which is where the factoring stops

Answer

no, a four-term polynomial has to be split into three pairs before it can group

Question 3

Factor 8t^3 + 20t^2 - 6t - 15 completely.

Accepted Answer

(2t + 5)(4t^2 - 3)

Answer

4t^2(2t + 5) - 3(2t - 5)

Answer

4t^2(2t + 5) - 3(2t + 5)

Answer

(2t + 5)(4t^2 + 3)

Question 4

Factor 25w^3 + 40w^2 + 10w + 16 completely.

Accepted Answer

(5w + 8)(5w^2 + 2)

Answer

5w^2(5w + 8) + 2(5w + 8)

Answer

(5w + 2)(5w^2 + 8)

Answer

(5w + 8)(5w^2 - 2)

Question 5

Factor 15y^3 + 10y^2 - 12y - 8 completely.

Accepted Answer

(3y + 2)(5y^2 - 4)

Answer

5y^2(3y + 2) - 4(3y - 2)

Answer

5y^2(3y + 2) - 4(3y + 2)

Answer

(3y + 2)(5y^2 + 4)

Question 6

Grouping 6h^3 + 27h^2 + 16h + 72 in pairs gives 3h^2(2h + 9) + 8(2h + 9). Which binomial factor do the two parts share?

Accepted Answer

(2h + 9)

Answer

(3h^2 + 8)

Answer

(3h + 8)

Answer

(2h - 9)

Question 7

Grouping 4r^3 + 3r^2 + 12r + 9 in pairs gives r^2(4r + 3) + 3(4r + 3). Which binomial factor do the two parts share?

Accepted Answer

(4r + 3)

Answer

(r^2 + 3)

Answer

(r + 3)

Answer

(4r - 3)

Question 8

Factor 15h^3 + 25h^2 + 6h + 10 completely.

Accepted Answer

(3h + 5)(5h^2 + 2)

Answer

5h^2(3h + 5) + 2(3h + 5)

Answer

(5h + 2)(3h^2 + 5)

Answer

(3h + 5)(5h^2 - 2)

Question 9

Factor 3h^3 + 12h^2 - 5h - 20 completely.

Accepted Answer

(h + 4)(3h^2 - 5)

Answer

3h^2(h + 4) - 5(h - 4)

Answer

3h^2(h + 4) - 5(h + 4)

Answer

(h + 4)(3h^2 + 5)

Question 10

Factor 15y^3 + 24y^2 - 25y - 40 completely.

Accepted Answer

(5y + 8)(3y^2 - 5)

Answer

3y^2(5y + 8) - 5(5y - 8)

Answer

3y^2(5y + 8) - 5(5y + 8)

Answer

(5y + 8)(3y^2 + 5)

Question 11

Factor 8y^3 + 12y^2 - 10y - 15 completely.

Accepted Answer

(2y + 3)(4y^2 - 5)

Answer

4y^2(2y + 3) - 5(2y - 3)

Answer

4y^2(2y + 3) - 5(2y + 3)

Answer

(2y + 3)(4y^2 + 5)

Question 12

Factor 15x^3 + 20x^2 + 24x + 32 completely.

Accepted Answer

(3x + 4)(5x^2 + 8)

Answer

5x^2(3x + 4) + 8(3x + 4)

Answer

(5x + 8)(3x^2 + 4)

Answer

(3x + 4)(5x^2 - 8)

Question 13

Factor 2a^3 + 4a^2 - 9a - 18 completely.

Accepted Answer

(a + 2)(2a^2 - 9)

Answer

2a^2(a + 2) - 9(a - 2)

Answer

2a^2(a + 2) - 9(a + 2)

Answer

(a + 2)(2a^2 + 9)

Question 14

Does y^3 + 5y^2 + 4y + 20 factor by grouping?

Accepted Answer

yes, both pairs leave (y + 5), so it factors as (y + 5)(y^2 + 4)

Answer

no, the two pairs leave different binomials, so this one does not group at all

Answer

yes, but only as far as y^2(y + 5) + 4(y + 5), which is where the factoring stops

Answer

no, a four-term polynomial has to be split into three pairs before it can group

Question 15

Does 4n^3 + 18n^2 + 18n + 82 factor by grouping?

Accepted Answer

no, the two pairs leave different binomials, so this one does not group at all

Answer

yes, both pairs leave (2n + 9), so it factors as (2n + 9)(2n^2 + 9)

Answer

yes, but only as far as 2n^2(2n + 9) + 9(2n + 9), which is where the factoring stops

Answer

yes, any polynomial with four terms can be factored by grouping them in pairs

Question 16

Grouping 2t^3 + 3t^2 + 10t + 15 in pairs gives t^2(2t + 3) + 5(2t + 3). Which binomial factor do the two parts share?

Accepted Answer

(2t + 3)

Answer

(t^2 + 5)

Answer

(t + 5)

Answer

(2t - 3)

Question 17

Factor a^3 + 4a^2 + 9a + 36 completely.

Accepted Answer

(a + 4)(a^2 + 9)

Answer

a^2(a + 4) + 9(a + 4)

Answer

(a + 9)(a^2 + 4)

Answer

(a + 4)(a^2 - 9)

Question 18

Factor 6a^3 + 21a^2 + 16a + 56 completely.

Accepted Answer

(2a + 7)(3a^2 + 8)

Answer

3a^2(2a + 7) + 8(2a + 7)

Answer

(3a + 8)(2a^2 + 7)

Answer

(2a + 7)(3a^2 - 8)

Question 19

Factor 25n^3 + 20n^2 + 30n + 24 completely.

Accepted Answer

(5n + 4)(5n^2 + 6)

Answer

5n^2(5n + 4) + 6(5n + 4)

Answer

(5n + 6)(5n^2 + 4)

Answer

(5n + 4)(5n^2 - 6)

Question 20

Factor 4r^3 + 3r^2 + 28r + 21 completely.

Accepted Answer

(4r + 3)(r^2 + 7)

Answer

r^2(4r + 3) + 7(4r + 3)

Answer

(r + 7)(4r^2 + 3)

Answer

(4r + 3)(r^2 - 7)

Question 21

Factor w^3 + 9w^2 - 3w - 27 completely.

Accepted Answer

(w + 9)(w^2 - 3)

Answer

w^2(w + 9) - 3(w - 9)

Answer

w^2(w + 9) - 3(w + 9)

Answer

(w + 9)(w^2 + 3)

Question 22

Factor 20h^3 + 15h^2 + 8h + 6 completely.

Accepted Answer

(4h + 3)(5h^2 + 2)

Answer

5h^2(4h + 3) + 2(4h + 3)

Answer

(5h + 2)(4h^2 + 3)

Answer

(4h + 3)(5h^2 - 2)

Question 23

Factor m^3 + 3m^2 + 5m + 15 completely.

Accepted Answer

(m + 3)(m^2 + 5)

Answer

m^2(m + 3) + 5(m + 3)

Answer

(m + 5)(m^2 + 3)

Answer

(m + 3)(m^2 - 5)

Question 24

Factor 2t^3 + 5t^2 + 14t + 35 completely.

Accepted Answer

(2t + 5)(t^2 + 7)

Answer

t^2(2t + 5) + 7(2t + 5)

Answer

(t + 7)(2t^2 + 5)

Answer

(2t + 5)(t^2 - 7)

Question 25

Factor 20x^3 + 15x^2 + 28x + 21 completely.

Accepted Answer

(4x + 3)(5x^2 + 7)

Answer

5x^2(4x + 3) + 7(4x + 3)

Answer

(5x + 7)(4x^2 + 3)

Answer

(4x + 3)(5x^2 - 7)

Question 26

Grouping 5a^3 + 35a^2 + 9a + 63 in pairs gives 5a^2(a + 7) + 9(a + 7). Which binomial factor do the two parts share?

Accepted Answer

(a + 7)

Answer

(5a^2 + 9)

Answer

(5a + 9)

Answer

(a - 7)

Question 27

Factor 6w^3 + 8w^2 - 15w - 20 completely.

Accepted Answer

(3w + 4)(2w^2 - 5)

Answer

2w^2(3w + 4) - 5(3w - 4)

Answer

2w^2(3w + 4) - 5(3w + 4)

Answer

(3w + 4)(2w^2 + 5)

Question 28

Grouping k^3 + 3k^2 + 6k + 18 in pairs gives k^2(k + 3) + 6(k + 3). Which binomial factor do the two parts share?

Accepted Answer

(k + 3)

Answer

(k^2 + 6)

Answer

(k + 6)

Answer

(k - 3)

Question 29

Factor 5h^3 + 10h^2 + 7h + 14 completely.

Accepted Answer

(h + 2)(5h^2 + 7)

Answer

5h^2(h + 2) + 7(h + 2)

Answer

(5h + 7)(h^2 + 2)

Answer

(h + 2)(5h^2 - 7)

Question 30

Factor 20r^3 + 8r^2 - 15r - 6 completely.

Accepted Answer

(5r + 2)(4r^2 - 3)

Answer

4r^2(5r + 2) - 3(5r - 2)

Answer

4r^2(5r + 2) - 3(5r + 2)

Answer

(5r + 2)(4r^2 + 3)

Question 31

Factor 4k^3 + 12k^2 - 7k - 21 completely.

Accepted Answer

(k + 3)(4k^2 - 7)

Answer

4k^2(k + 3) - 7(k - 3)

Answer

4k^2(k + 3) - 7(k + 3)

Answer

(k + 3)(4k^2 + 7)

Question 32

Factor n^3 + 2n^2 - 3n - 6 completely.

Accepted Answer

(n + 2)(n^2 - 3)

Answer

n^2(n + 2) - 3(n - 2)

Answer

n^2(n + 2) - 3(n + 2)

Answer

(n + 2)(n^2 + 3)

Question 33

Factor 5m^3 + 10m^2 + 2m + 4 completely.

Accepted Answer

(m + 2)(5m^2 + 2)

Answer

5m^2(m + 2) + 2(m + 2)

Answer

(5m + 2)(m^2 + 2)

Answer

(m + 2)(5m^2 - 2)

Question 34

Factor 6m^3 + 15m^2 + 14m + 35 completely.

Accepted Answer

(2m + 5)(3m^2 + 7)

Answer

3m^2(2m + 5) + 7(2m + 5)

Answer

(3m + 7)(2m^2 + 5)

Answer

(2m + 5)(3m^2 - 7)

Question 35

Factor p^3 + 4p^2 - 8p - 32 completely.

Accepted Answer

(p + 4)(p^2 - 8)

Answer

p^2(p + 4) - 8(p - 4)

Answer

p^2(p + 4) - 8(p + 4)

Answer

(p + 4)(p^2 + 8)

Question 36

Factor 9p^3 + 6p^2 + 6p + 4 completely.

Accepted Answer

(3p + 2)(3p^2 + 2)

Answer

3p^2(3p + 2) + 2(3p + 2)

Answer

(3p + 2)(3p^2 - 2)

Answer

(3p - 2)(3p^2 + 2)

Question 37

Does 4r^3 + 3r^2 + 12r + 9 factor by grouping?

Accepted Answer

yes, both pairs leave (4r + 3), so it factors as (4r + 3)(r^2 + 3)

Answer

no, the two pairs leave different binomials, so this one does not group at all

Answer

yes, but only as far as r^2(4r + 3) + 3(4r + 3), which is where the factoring stops

Answer

no, a four-term polynomial has to be split into three pairs before it can group

Question 38

Grouping 3a^3 + 2a^2 + 15a + 10 in pairs gives a^2(3a + 2) + 5(3a + 2). Which binomial factor do the two parts share?

Accepted Answer

(3a + 2)

Answer

(a^2 + 5)

Answer

(a + 5)

Answer

(3a - 2)

Question 39

Factor 15a^3 + 6a^2 + 25a + 10 completely.

Accepted Answer

(5a + 2)(3a^2 + 5)

Answer

3a^2(5a + 2) + 5(5a + 2)

Answer

(3a + 5)(5a^2 + 2)

Answer

(5a + 2)(3a^2 - 5)

Question 40

Factor 12y^3 + 32y^2 + 15y + 40 completely.

Accepted Answer

(3y + 8)(4y^2 + 5)

Answer

4y^2(3y + 8) + 5(3y + 8)

Answer

(4y + 5)(3y^2 + 8)

Answer

(3y + 8)(4y^2 - 5)

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Factoring by Grouping

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Factoring by Grouping

Ways to Play

Device Games: Students play on their phones

Blast Go

Zombie Escape

Avalanche

Maze Blast

Cave Climb

Blast Out

Mystery Auction

Tour Guide

Food Truck Rush

Tug of War

Alien Blast

Glacier Rush

Board Games: Teacher-led, projected to the class

Like what you see?

More like this

Bug Blast

BlastRace

BlastQuiz

BlastBoard

Seek & Destroy

Mystery Blast

Checkers

Texas Hold 'Em

Da Bomb

Castle Siege

Hot Potato

Market Mayhem

King of the Hill

Bank Heist

Top Corner