Factoring Trinomials Using the Slide and Divide Method Practice

Factoring a Trinomial Using the Slide and Divide Method

At this point, we know how to factor a trinomial using both the AC and Reverse FOIL methods. Each method has its pros and cons. Reverse FOIL is typically only used when the number of trials is kept to a minimum; otherwise, using the AC Method is normally the way to go.

Here, we will introduce an alternative approach known as the Slide and Divide Method that can sometimes be a bit faster. It won't always work out to be the case, but many students really like this approach.

Slide and Divide Method for Factoring a Trinomial

1. Factor out the GCF if it isn't 1
   • If the leading coefficient is negative, factoring out the -GCF is helpful
2. Write the trinomial in standard form
   • ax² + bx + c
3. Slide the leading coefficient "a" to the constant term "c"
   • Replace the constant term "c" with the product "ac"
   • Change the leading coefficient "a" into 1
4. Factor the new trinomial using the method for a = 1
   • Find two integers whose sum is the middle coefficient and whose product is the constant term
   • Use those two integers, p and q, to complete each binomial
     • (x + p)(x + q)
5. Divide both integers p and q by the original "a" value
6. Simplify each fraction
   • If a denominator is not 1, slide it in front of the x in that binomial
7. If no such integers p and q can be found, the polynomial is prime

Let's look at an example.

Example #1: Factor each trinomial. $$4x^2 + 39x + 27$$

1) The GCF of 4x², 39x, and 27 is 1.

2) We will match up the a, b, and c values. For reference:

$$ax^2 + bx + c$$ $$4x^2 + 39x + 27$$ $$a = 4, b = 39, c = 27$$ $$ac = 4 \cdot 27 = 108$$

3) Slide the leading coefficient "a" (4) to the constant term "c" (27). This means we will replace the leading coefficient "a" with 1, and replace the constant term "c" with "ac" (108).

$$x^2 + 39x + 108$$

4) Factor the new trinomial using the method for a = 1. This means we are going to find two integers whose sum is 39 and whose product is 108. Since all signs are positive, we can just think about positive factor pairs of 108.

Factor 1	Factor 2	Sum
1	108	109
2	54	56
3	36	39
4	27	31
6	18	24
9	12	21

From the table above, we can see the required integers are 3 and 36. We can use this to factor our trinomial.

$$(x + 3)(x + 36)$$

5) At this point, we will divide the integers (3 and 36) placed in the final position of each binomial by the original a-value (4).

$$\left(x + \frac{3}{4}\right)\left(x + \frac{36}{4}\right)$$

6) Simplify each fraction:
3/4 isn't going to simplify, but 36/4 = 9.

$$\left(x + \frac{3}{4}\right)\left(x + 9\right)$$

7) In the leftmost binomial, we have a denominator of 4. Slide this in front of the x.

$$(4x + 3)(x + 9)$$ We can check with FOIL: $$\text{F:} \, 4x \cdot x = 4x^2$$ $$\text{O:} \, 4x \cdot 9 = 36x$$ $$\text{I:} \, 3 \cdot x = 3x$$ $$\text{L:} \, 3 \cdot 9 = 27$$ $$\text{O + I:} \, 36x + 3x = 39x$$ $$(4x + 3)(x + 9) = 4x^2 + 39x + 27 \, ✓$$

When you have a common factor involved in the trinomial, it must be taken out first; otherwise, you will get the wrong answer. Let's look at an example.

Example #2: Factor each trinomial. $$12x^2 - 51x - 45$$

1) The GCF of 12x², -51x, and -45 is 3.

$$12x^2 - 51x - 45 = 3(4x^2 - 17x - 15)$$

2) We will match up the a, b, and c values. For reference:

$$ax^2 + bx + c$$ Don't worry about the 3 for right now, just focus on the trinomial inside of the parentheses. At the end of the problem, we will bring the 3 back. $$4x^2 - 17x - 15$$ $$a = 4, b = -17, c = -15$$ $$ac = 4 \cdot -15 = -60$$

3) Slide the leading coefficient "a" (4) to the constant term "c" (-15). This means we will replace the leading coefficient "a" with 1, and replace the constant term "c" with "ac" (-60).

$$x^2 - 17x - 60$$

4) Factor the new trinomial using the method for a = 1. This means we are going to find two integers whose sum is -17 and whose product is -60. Since we have a negative product and a negative sum, we are looking for mixed signs where the larger in terms of absolute value is negative.

Factor 1	Factor 2	Sum
1	-60	-59
2	-30	-28
3	-20	-17
4	-15	-11
5	-12	-7
6	-10	-4

From the table above, we can see the required integers are 3 and -20. We can use this to factor our trinomial.

$$(x + 3)(x - 20)$$

5) At this point, we will divide the integers (3 and -20) placed in the final position of each binomial by the original a-value (4).

$$\left(x + \frac{3}{4}\right)\left(x + \frac{-20}{4}\right)$$

6) Simplify each fraction:
3/4 isn't going to simplify, but -20/4 = -5.

$$\left(x + \frac{3}{4}\right)\left(x - 5\right)$$

7) In the leftmost binomial, we have a denominator of 4. Slide this in front of the x.

$$(4x + 3)(x - 5)$$ Now we can add our GCF (3) back in. Make sure to include this in your final answer; otherwise, it will be marked as incorrect. $$3(4x + 3)(x - 5)$$ We can check with FOIL: $$\text{F:} \, 4x \cdot x = 4x^2$$ $$\text{O:} \, 4x \cdot -5 = -20x$$ $$\text{I:} \, 3 \cdot x = 3x$$ $$\text{L:} \, 3 \cdot -5 = -15$$ $$\text{O + I:} \, -20x + 3x = -17x$$ $$(4x + 3)(x - 5) = 4x^2 - 17x - 15$$ Multiply by the GCF (3): $$3(4x + 3)(x - 5)$$ $$= 3(4x^2 - 17x - 15)$$ $$= 12x^2 - 51x - 45 \, ✓$$

Why Does Slide and Divide Work?

Let's work through a simple example to show the reasoning for slide and divide.

$$6x^2 + 13x + 6$$ First, let's do slide and divide. Afterwards, we can compare our results. $$x^2 + 13x + 36$$ $$(x + 4)(x + 9)$$ $$\left(x + \frac{4}{6}\right)\left(x + \frac{9}{6}\right)$$ $$\left(x + \frac{2}{3}\right)\left(x + \frac{3}{2}\right)$$ $$(3x + 2)(2x + 3)$$ Now let's do this with a substitution technique. To make the math easier to follow, we will set our trinomial equal to y. $$y = 6x^2 + 13x + 6$$ Now we will introduce a simple substitution. Let x = z/6 $$y = 6\left(\frac{z}{6}\right)^2 + 13 \cdot \frac{z}{6} + 6$$ $$y = 6 \cdot \frac{z^2}{36} + 13 \cdot \frac{z}{6} + 6$$ $$y = \frac{z^2}{6} + 13 \cdot \frac{z}{6} + 6$$ Multiply both sides by 6: $$6y = z^2 + 13z + 36$$ So this sets up the slide. Notice that the coefficient of z² is 1, and the c is replaced with ac. At this point, we can factor the right side. $$6y = (z + 4)(z + 9)$$ Since we let x = z/6, we can solve this for z by multiplying both sides by 6. $$z = 6x$$ Now we can replace each z with 6x. $$6y = (6x + 4)(6x + 9)$$ Pull out the GCF from each: $$6y = 2(3x + 2) \cdot 3(2x + 3)$$ $$6y = 6(3x + 2)(2x + 3)$$ Divide both sides by 6 to solve for y: $$y = (3x + 2)(2x + 3)$$ Note: Pulling out the GCF doesn't match up entirely with our slide and divide method, but it's the most efficient path at that point. Here is what we can do to make it match. $$6y = (6x + 4)(6x + 9)$$ Factor out 6 from each binomial: $$6y = 6\left(x + \frac{2}{3}\right) \cdot 6\left(x + \frac{3}{2}\right)$$ $$6y = 36\left(x + \frac{2}{3}\right)\left(x + \frac{3}{2}\right)$$ Divide each side by 6: $$y = 6\left(x + \frac{2}{3}\right)\left(x + \frac{3}{2}\right)$$ Rewrite 6 as 2 × 3 and "slide" each into the appropriate binomial to cancel the denominator inside. $$y = 2 \cdot 3 \left(x + \frac{2}{3}\right)\left(x + \frac{3}{2}\right)$$ $$y = 3\left(x + \frac{2}{3}\right) \cdot 2\left(x + \frac{3}{2}\right)$$ $$y = (3x + 2)(2x + 3)$$ You can see this matches the faster approach where each denominator just slides to become the coefficient of x in its binomial.