Solved 5.5 Worksheet - Factoring ax2+bx+c MPM2D Jensen 2. | Chegg.com - Free Printable

Let’s solve each problem step by step. We’re factoring trinomials of the form ax² + bx + c. The goal is to write them as two binomials: ( __ )( __ ).

We’ll use the “AC method” or trial-and-error with factors — whichever works best for each one.

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Problem 2a:

2x² + 7x + 5

Multiply a·c = 2·5 = 10
Find two numbers that multiply to 10 and add to 7 → 2 and 5

Rewrite middle term:
2x² + 2x + 5x + 5

Group:
(2x² + 2x) + (5x + 5)
= 2x(x + 1) + 5(x + 1)
= (2x + 5)(x + 1)

✔ Check: (2x+5)(x+1) = 2x² + 2x + 5x + 5 = 2x² + 7x + 5 ✔️

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Problem 2b:

6y² + 19y + 8

a·c = 6·8 = 48
Need two numbers that multiply to 48, add to 19 → 3 and 16

Rewrite:
6y² + 3y + 16y + 8

Group:
(6y² + 3y) + (16y + 8)
= 3y(2y + 1) + 8(2y + 1)
= (3y + 8)(2y + 1)

✔ Check: (3y+8)(2y+1) = 6y² + 3y + 16y + 8 = 6y² + 19y + 8 ✔️

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Problem 2c:

4k² + 15k + 9

a·c = 4·9 = 36
Numbers that multiply to 36, add to 15 → 3 and 12

Rewrite:
4k² + 3k + 12k + 9

Group:
(4k² + 3k) + (12k + 9)
= k(4k + 3) + 3(4k + 3)
= (k + 3)(4k + 3)

✔ Check: (k+3)(4k+3) = 4k² + 3k + 12k + 9 = 4k² + 15k + 9 ✔️

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Problem 2d:

3m² + 10m + 8

a·c = 3·8 = 24
Numbers that multiply to 24, add to 10 → 4 and 6

Rewrite:
3m² + 4m + 6m + 8

Group:
(3m² + 4m) + (6m + 8)
= m(3m + 4) + 2(3m + 4)
= (m + 2)(3m + 4)

✔ Check: (m+2)(3m+4) = 3m² + 4m + 6m + 8 = 3m² + 10m + 8 ✔️

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Problem 2e:

10w² + 15w + 3

a·c = 10·3 = 30
Need two numbers that multiply to 30, add to 15 → Let’s list factor pairs of 30:

1 & 30 → sum 31
2 & 15 → sum 17
3 & 10 → sum 13
5 & 6 → sum 11 ← none add to 15!

Wait — maybe it doesn’t factor? Let’s check discriminant: b² - 4ac = 15² - 4·10·3 = 225 - 120 = 105 → not a perfect square → cannot be factored over integers

So answer: Not factorable (or leave as is)

But let’s double-check if we missed something... No, no integer pair multiplies to 30 and adds to 15. So yes — not factorable.

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Problem 2f:

12q² + 17q + 6

a·c = 12·6 = 72
Need two numbers that multiply to 72, add to 17 → Try: 8 and 9 → 8×9=72, 8+9=17 ✔

Rewrite:
12q² + 8q + 9q + 6

Group:
(12q² + 8q) + (9q + 6)
= 4q(3q + 2) + 3(3q + 2)
= (4q + 3)(3q + 2)

✔ Check: (4q+3)(3q+2) = 12q² + 8q + 9q + 6 = 12q² + 17q + 6 ✔️

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Problem 3a:

4x² - 11x + 6

a·c = 4·6 = 24
Need two numbers that multiply to 24, add to -11 → both negative: -3 and -8 → (-3)(-8)=24, -3+-8=-11 ✔

Rewrite:
4x² - 3x - 8x + 6

Group:
(4x² - 3x) + (-8x + 6)
= x(4x - 3) -2(4x - 3)
= (x - 2)(4x - 3)

✔ Check: (x-2)(4x-3) = 4x² -3x -8x +6 = 4x² -11x +6 ✔️

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Problem 3b:

5n² - 11n + 6

a·c = 5·6 = 30
Need two numbers that multiply to 30, add to -11 → -5 and -6 → (-5)(-6)=30, -5+-6=-11 ✔

Rewrite:
5n² - 5n - 6n + 6

Group:
(5n² - 5n) + (-6n + 6)
= 5n(n - 1) -6(n - 1)
= (5n - 6)(n - 1)

✔ Check: (5n-6)(n-1) = 5n² -5n -6n +6 = 5n² -11n +6 ✔️

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Problem 3e:

9b² - 24b + 7

a·c = 9·7 = 63
Need two numbers that multiply to 63, add to -24 → both negative: -3 and -21 → (-3)(-21)=63, -3+-21=-24 ✔

Rewrite:
9b² - 3b - 21b + 7

Group:
(9b² - 3b) + (-21b + 7)
= 3b(3b - 1) -7(3b - 1)
= (3b - 7)(3b - 1)

✔ Check: (3b-7)(3b-1) = 9b² -3b -21b +7 = 9b² -24b +7 ✔️

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Final Answer:

2a) (2x + 5)(x + 1)
2b) (3y + 8)(2y + 1)
2c) (k + 3)(4k + 3)
2d) (m + 2)(3m + 4)
2e) Not factorable
2f) (4q + 3)(3q + 2)

3a) (x - 2)(4x - 3)
3b) (5n - 6)(n - 1)
3e) (3b - 7)(3b - 1)