Key Concepts
- Factor a quadratic trinomial when a is not equal to 1
- Factor out a Common Factor
- Understand factoring by grouping
- Factor a trinomial using substitution
Factoring a trinomial when a is not equal to 1
Factor out a Common Factor
What is the factored form of 3x3 + 15x2 – 18x?
Before factoring the trinomial into two binomials, look for any common factors that you can factor out.
So, 3x3 + 15x2 – 18x = 3x(x2 + 5x – 6).
Then factor the resulting trinomial, x2 + 5x – 6.
The factored form of x2 + 5x – 6 is (x – 1)(x + 6), so the factored form of 3x3 + 15x2 – 18x is 3x(x – 1)(x + 6).
Understand factoring by grouping
- If ax2 + bx + c is a product of binomials, how are the values of a, b and c related? Consider the product (3x + 4)(2x + 1).
The product is 6x2 + 11x + 4. Notice that ac = (6)(4) or (3)(2)(4)(1), which is the product of all of the coefficients and constants from (3x + 4)(2x + 1).
In the middle terms, the coefficients of the x-terms, 3 and 8, add to form b = 11. They are composed of the pairs of the coefficients and constants from the original product; 3 = (3)(1) and 8 = (4)(2).
If ax2 + bx + c is the product of the binomials, there is a pair of the factors of ac that have a sum of b.
- How can you factor ax2 + bx + c by grouping?
Consider the trinomial 6x2 + 11x + 4, a = 6 and c = 4, so ac = 24.
Find the factor pair of 24 with a sum of 11
The factored form of 6x2 + 11x + 4 is (3x + 4)(2x + 1).
Check. (3x + 4)(2x + 1) = 6x2 + 3x + 8x + 4 = 6x2 + 11x + 4
- Factoring a trinomial using substitution method
How can you use substitution to help you factor ax2 + bx + c as the product of two binomials?
Consider the trinomial 3x2 – 2x – 8.
Step 1. Multiply ax2 + bx + c by a to transform x2 into (ax)2.
Step 2. Replace ax with a single variable. Let p = ax.
= p2 – 2p – 24
Step 3. Factor the trinomial.
= (p – 6)(p + 4)
Step 4.
Substitute ax back into the product. Remember p = 3x. Factor out the common factors if there are any.
Step 5.
Since you started by multiplying the trinomial by a, you must divide by a to get a product that is equivalent to original trinomial.
The factored form of 3x2 – 2x – 8 is (x – 2)(3x + 4). In general, you can use substitution to help transform ax2 + bx + c with a not equal to 1 to a simpler case in which a = 1, factor it, and then transform it back to an equivalent factored form.
Questions
Question 1
Write the factored form of each trinomial.
1. 5x2 – 35x + 50
Take out 5.
5(x2 – 7x + 10)
Find the factors of x2 – 7x + 10
Factors of 10 are -5 and -2.
-5 + (-2) = -7
So, (x – 5)(x – 2)
Ans: 5(x – 5)(x – 2)
2. 6x3 + 30x2 + 24x
Take out 6x.
6x(x2 + 5x + 4)
Find the factors of x2 + 5x + 4
Factors of 4 are 4 and 1.
4 + 1 = 5
So, (x + 4)(x + 1)
Ans: 6x(x + 4)(x + 1)
3. 10x2 + 17x + 3
a = 10, b = 17, c = 3
a × c = 10 × 3 = 30
Factors of 30 are 15 and 2.
And 15 + 2 = 17
10x2 + 15x + 2x + 3
= 5x(2x + 3) + 1(2x + 3)
= (5x + 1)(2x + 3)
4. 2x2 – x – 6
a = 2, b = -1, c = -6
a × c = -12
Factors of -12 are -4 and 3.
And (-4) + 3 = -1
2x2 – 4x + 3x – 6
= 2x(x – 2) + 3(x – 2)
= (2x + 3)(x – 2)
5. 10x2 + 3x – 1
a = 10, b = 3, c = -1
a*c = -10
Factors of -10 are 5 and -2.
And 5 + (-2) = 3
10x2 + 5x – 2x – 1
= 5x(2x + 1) + (-1)(2x + 1)
= (5x – 1)(2x + 1)
Question 2
A photographer is placing photos in a mat for a gallery show. Each mat she uses is x in. wide on each side. The total area of each photo and mat is shown below.
- Factor the total area to find the possible dimensions of a photo and mat.
- What are the dimensions of the photos in terms of x?
Solution:
1. Total area of the photo and the mat = 4x^2 + 36x + 80
Let’s factor this trinomial using the substitution method.
(2x) 2 + 18(2x) + 80
Let 2x = p.
The trinomial becomes p2 + 18p + 80.
Factors of 80 are 10 and 8.
And 10 + 8 = 18
(p + 10)(p + 8)
As p = 2x,
4x2 + 36x + 80 = (2x + 10)(2x + 8)
2. The dimensions of the photo and the mat combined are 2x + 10 in. by 2x + 8 in.
To find the dimension of just the photo, subtract 2x from both length and width.
L = (2x + 10) – 2x = 10 in.
W = (2x + 8) – 2x = 8 in.
The dimensions of each photo are 10 in. by 8 in. In terms of x, it is 10x0 in. by 8x0 in. (independent of the value of x).
Key Concepts Covered
Exercise
Factor the following trinomials:
- 8x2 – 10x – 3
- 12x2+ 16x + 5
- 16x3 + 32x2 + 12x
- 21x2 – 35x – 14
- 16x2 + 22x – 3
- 9x2 + 46x + 5
- –6x2 – 25x – 25
- 5x2 – 4xy – y2
- 16x2 + 60x – 100
- 6x2 + 5x – 6
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