Factoring when the leading coefficient is not one
When x squared carries a coefficient, two numbers that multiply to the constant no longer give the middle term. What do you look for instead?
Where you are 3 in Factoring trinomials
When you reach for it
You meet a trinomial ax^2 + bx + c with integer coefficients and a not 1, and you want it as a product of two binomials. When a = 1, the shortcut is to find two numbers that multiply to c and add to b. Try it on 2x^2 + 11x + 12. The factor pairs of 12 are 1, 12 and 2, 6 and 3, 4, with sums 13, 8 and 7. None is 11, yet the trinomial does factor, as multiplying out shows:
The middle term is 2 \cdot 4 + 3 \cdot 1, not 3 + 4. The 2 in front of x multiplies the 4 before anything is added, so the leading coefficient is mixed into the middle term, and a search among the factors of c alone cannot see it.
Factored form is what later work needs: reading the zeros of a quadratic and solving equations with the zero product property both start from it. This lesson is about the factoring itself; solving comes later.
The procedure
This lesson runs over two days. Day 1 is the main method, splitting the middle term. Day 2 adds a second route and the other shapes a trinomial can take.
Day 1: splitting the middle term
Draw the product (2x + 3)(x + 4) as the area of a rectangle.
The four pieces are 2x^2, 8x, 3x and 12. The middle term 11x is not one piece but two, 8x and 3x, and there is a pattern in their coefficients: 8 \cdot 3 = 24, and the corner pieces give 2 \cdot 12 = 24 as well. So to rebuild the four pieces from 2x^2 + 11x + 12, look for two numbers whose product is ac = 2 \cdot 12 = 24 and whose sum is b = 11.
The pairs for 24 are 1, 24 and 2, 12 and 3, 8, and 3 + 8 = 11. Rewrite 11x as 8x + 3x, which changes nothing since 8x + 3x = 11x:
Four terms now, and four terms are what Factoring by grouping handles. Take 2x out of the first pair and 3 out of the second:
The last step is the distributive property used backwards, with the shared bracket (x + 4) as the common factor.
Why the product is ac. Expand a general product of two binomials:
So a = pr, b = ps + qr and c = qs. The two middle coefficients ps and qr add to b, and by the commutative and associative properties of multiplication their product is
So whenever ax^2 + bx + c factors into binomials with integer coefficients, two integers exist with product ac and sum b, namely ps and qr. Finding them recovers the four pieces of the picture. With a = 1 this is the old shortcut, since then ac = c.
Why grouping then always works. Say m and n have mn = ac and m + n = b; both are non-zero, because their product ac is not. Then mn = ac says the ratio a : m equals n : c, so once the largest common factor of each pair is taken out, with the sign chosen to match, the brackets in ax^2 + mx and nx + c are the same. Which of m and n comes first does not matter either.
Signs. The sign of ac and of b narrow the search before you list anything.
| ac | b | the two numbers |
|---|---|---|
| positive | positive | both positive |
| positive | negative | both negative |
| negative | either | opposite signs, the one larger in size has the sign of b |
Two numbers of the same sign multiply to a positive and add to something with their shared sign. Two of opposite signs multiply to a negative, and their sum takes the sign of the larger one.
Day 2: trial and check, and other shapes
Trial and check. A second route goes straight for the brackets. In (px + q)(rx + s), the first terms must multiply to a and the last terms to c, so try factor pairs of a for p, r and factor pairs of c for q, s, and keep the arrangement whose outer and inner products add to b, that is ps + qr = b. Two facts cut the list. With p and r positive, the first two rows of the sign table hold for q and s, with c in place of ac; when c is negative, q and s simply have opposite signs. And if the trinomial has no common factor, neither bracket has one, because a factor common to one bracket would divide every term of the product. So 2x + 4 can never be a bracket of 6x^2 + 7x + 2, say.
When each route is shorter. Trial and check is quick when a and c have few factor pairs: in 5x^2 - 13x - 6 the first terms can only be 5x and x. When a and c both have many factors it is slow. For 12t^2 + 25t + 12 there are 3 ways to pick the first terms and 6 ordered ways to pick the last ones, 18 arrangements, while splitting the middle term lists the pairs of 144 once and stops at 9 and 16. Both routes rest on the same equation, ps + qr = b.
A negative leading coefficient. Take out -1 first. Every sign inside flips, and the trinomial in the bracket has a positive leading coefficient, so the methods above apply. Example 6 does this.
Two variables. In 10x^2 - xy - 2y^2 every term has degree two. The coefficients 10, -1 and -2 behave exactly as in one variable, and y rides along with the last term of each bracket. Example 7 works it.
When it does not factor. Take 3x^2 + 10x + 4. Here ac = 12 and b = 10. Since both are positive the numbers would both be positive, and the complete list of positive pairs for 12 gives the sums 1 + 12 = 13, 2 + 6 = 8 and 3 + 4 = 7. None is 10. If the trinomial factored as (px + q)(rx + s) with integers, ps and qr would be such a pair, so it does not factor over the integers. Such a trinomial is called prime. The list is the reason, which is why it has to be complete.
Checking. Multiplying the factors back out takes a few seconds and catches every slip, a wrong sign above all. Each example below ends with it.
Worked examples
Day 1: both numbers positive
Factor 3x^2 + 17x + 10.
SolutionHere ac = 3 \cdot 10 = 30 and b = 17. Both are positive, so look for two positive numbers. The pairs for 30 are 1, 30 with sum 31, then 2, 15 with sum 17. Stop there. Split 17x as 2x + 15x and group:
Check. (x + 5)(3x + 2) = 3x^2 + 2x + 15x + 10, which is 3x^2 + 17x + 10.
Day 1: a negative middle term
Factor 4x^2 - 16x + 15.
SolutionHere ac = 60 is positive and b = -16 is negative, so both numbers are negative. The pairs for 60 are 1, 60 and 2, 30 and 3, 20 and 4, 15 and 5, 12 and 6, 10, with sums 61, 32, 23, 19, 17 and 16. So the numbers are -6 and -10. Split and group:
In the second pair, -5 comes out, not 5, so that the bracket left behind is (2x - 3) to match the first one: -5(2x - 3) = -10x + 15.
Check. (2x - 5)(2x - 3) = 4x^2 - 6x - 10x + 15, which is 4x^2 - 16x + 15.
Day 1: a negative constant
Factor 6x^2 + x - 12.
SolutionHere ac = -72, so the two numbers have opposite signs, and since b = 1 is positive the positive one is larger in size, by exactly 1. Two numbers that multiply to 72 and differ by 1 are 8 and 9, so the pair is 9 and -8:
Check. (3x - 4)(2x + 3) = 6x^2 + 9x - 8x - 12, which is 6x^2 + x - 12.
Day 1: the sides of a solar-panel array
A rectangular solar-panel array has area 10x^2 + 23x + 12 square meters, and each side is a length of the form px + q meters with positive integers p and q. Find the sides, and the array’s size when x = 2.
Area is length times width, so the sides are the factors. Here ac = 120 and b = 23, both positive. The pairs for 120 run 1, 120 and 2, 60 and 3, 40 and 4, 30 and 5, 24 and 6, 20, with sums 121, 62, 43, 34, 29, 26, and then 8, 15 with sum 23:
The sides are 2x + 3 and 5x + 4 meters. When x = 2 they are 7 and 14 meters.
Check. 7 \cdot 14 = 98, and 10 \cdot 4 + 23 \cdot 2 + 12 = 98.
Day 2: trial and check
Factor 5x^2 - 13x - 6 by trial and check.
SolutionThe first terms can only be 5x and x, so the factors are (5x + q)(x + s) with qs = -6. The outer and inner products add to 5s + q, which must be -13. As c is negative, q and s have opposite signs.
| q | s | 5s + q |
|---|---|---|
| 1 | -6 | -29 |
| -1 | 6 | 29 |
| 6 | -1 | 1 |
| -6 | 1 | -1 |
| 2 | -3 | -13 |
The fifth row works, so 5x^2 - 13x - 6 = (5x + 2)(x - 3).
Check. (5x + 2)(x - 3) = 5x^2 - 15x + 2x - 6, which is 5x^2 - 13x - 6. Splitting the middle term finds the same -15 and 2, from ac = -30; here both routes are short.
Day 2: a negative leading coefficient
Factor -4x^2 + 4x + 15.
SolutionTake out -1, which flips every sign inside:
Inside, ac = -60 and b = -4, so the numbers have opposite signs and the negative one is larger in size by 4. Since 6 \cdot 10 = 60, the pair is 6 and -10:
So -4x^2 + 4x + 15 = -(2x - 5)(2x + 3). The -1 may also go into a bracket, giving (5 - 2x)(2x + 3), the same product.
Check. (2x - 5)(2x + 3) = 4x^2 - 4x - 15, and multiplying by -1 gives -4x^2 + 4x + 15.
Day 2: two variables
Factor 10x^2 - xy - 2y^2.
SolutionRead the coefficients as a = 10, b = -1, c = -2. Then ac = -20, the signs are opposite, and the negative number is larger in size by 1: since 4 \cdot 5 = 20, the pair is 4 and -5. Split -xy as 4xy - 5xy:
Check. (2x - y)(5x + 2y) = 10x^2 + 4xy - 5xy - 2y^2, which is 10x^2 - xy - 2y^2.
Boundaries
Integer coefficients, and factors over the integers. The search lists integer pairs of ac. A trinomial with no such pair, such as 3x^2 + 10x + 4, is prime over the integers. That says nothing about whether it ever equals zero; it only says no factoring with integer coefficients exists.
A trinomial of degree two. The method needs a \ne 0, c \ne 0 and the shape ax^2 + bx + c. If c = 0, x is a common factor and taking it out is the whole job. Two variables work when every term has degree two, as in Example 7.
A common factor comes first. The pair search still works with one present, but the result is not fully factored: 6x^2 + 33x + 15 = 3(2x^2 + 11x + 5), which is 3(2x + 1)(x + 5). Taking it out first keeps the numbers small; Factoring completely and choosing a strategy treats this.
The leading coefficient should be positive. With a negative a, take out -1 first, as in Example 6.
Where this goes next
With a = 1 allowed as well, every trinomial that factors over the integers can now be factored, and Factoring completely and choosing a strategy puts this together with common factors and the special patterns into one strategy. Set the factored form equal to zero and the zero product property turns one quadratic equation into two linear ones, which is how the zeros of a quadratic are found when it factors. For the prime ones, completing the square still finds the solutions when there are any.
Exercises
Day 1: split the middle term
Factor.
- 2x^2 + 9x + 10
- 3x^2 + 10x + 8
- 5x^2 - 16x + 3
- 4y^2 - 4y - 3
- 6n^2 + 7n - 20
- 8x^2 - 2x - 15
- 12t^2 + 7t - 12
- 10x^2 - 29x + 10
Day 1: areas
- A rectangular rooftop deck has area 6x^2 + 17x + 12 square meters, and one side is 2x + 3 meters. Find the other side.
- A rectangular tapestry hangs on a wall. Its area is 15x^2 + 22x + 8 square feet, and each side is px + q feet with positive integers p and q. Find the sides, and the tapestry’s size when x = 4.
Day 1: reasoning
- Use the product (px + q)(rx + s) to explain why the two numbers in the split must multiply to ac and not to c.
- Aldwin factors 3x^2 + 17x + 10 by splitting 17x as 15x + 2x instead of 2x + 15x. Group his four terms, and explain why the order of the split cannot change the answer.
Day 2: trial and check
Factor by trial and check.
- 3x^2 + 16x + 5
- 7x^2 - 26x - 8
- 2x^2 - 13x + 21
Day 2: negative leading coefficients and two variables
Factor.
- -2x^2 + 5x + 12
- -6x^2 - 5x + 6
- 4x^2 + 11xy + 6y^2
- 10a^2 - 9ab - 9b^2
Day 2: factor, or show it does not
Factor, or show that it does not factor over the integers.
- 2x^2 + 5x + 4
- 3x^2 - 7x - 4
- 6x^2 - 5x - 4
- Find every integer b for which 2x^2 + bx + 3 factors over the integers.
Day 2: reasoning
- Explain why, when ax^2 + bx + c has no common factor, a bracket such as 2x + 6 can never be one of its factors.
Answers
- (2x + 5)(x + 2). ac = 20 = 4 \cdot 5 with 4 + 5 = 9: 2x(x + 2) + 5(x + 2).
- (x + 2)(3x + 4). ac = 24 = 4 \cdot 6 with 4 + 6 = 10: x(3x + 4) + 2(3x + 4).
- (x - 3)(5x - 1). ac = 15, both negative: -1 and -15. 5x^2 - x - 15x + 3 = x(5x - 1) - 3(5x - 1).
- (2y + 1)(2y - 3). ac = -12, pair -6 and 2: 2y(2y - 3) + 1(2y - 3).
- (3n - 4)(2n + 5). ac = -120, pair 15 and -8: 3n(2n + 5) - 4(2n + 5).
- (4x + 5)(2x - 3). ac = -120, pair -12 and 10: 4x(2x - 3) + 5(2x - 3).
- (4t - 3)(3t + 4). ac = -144, pair 16 and -9: 4t(3t + 4) - 3(3t + 4).
- (2x - 5)(5x - 2). ac = 100, pair -4 and -25: 2x(5x - 2) - 5(5x - 2).
- 3x + 4 meters. ac = 72, pair 8 and 9: 2x(3x + 4) + 3(3x + 4) = (2x + 3)(3x + 4).
- (5x + 4) by (3x + 2) feet, which is 24 by 14 feet, 336 square feet, when x = 4. ac = 120, pair 10 and 12: 5x(3x + 2) + 4(3x + 2).
- Expanding gives pr\,x^2 + (ps + qr)x + qs. The middle term is made of ps and qr, and (ps)(qr) = (pr)(qs) = ac. Their product involves p and r as well as q and s, so it is ac; it equals c only when a = pr = 1.
- 3x^2 + 15x + 2x + 10 = 3x(x + 5) + 2(x + 5), which is (3x + 2)(x + 5), the same factors. Addition is commutative, so both orders are the same four-term expression, and a factoring of one expression is a factoring of the other.
- (3x + 1)(x + 5). Outer and inner: 15x + x = 16x.
- (7x + 2)(x - 4). Outer and inner: -28x + 2x = -26x, and 2 \cdot (-4) = -8.
- (2x - 7)(x - 3). Both last terms negative since c > 0 and b < 0: -6x - 7x = -13x.
- -(2x + 3)(x - 4). -(2x^2 - 5x - 12); pair -8 and 3: 2x(x - 4) + 3(x - 4).
- -(3x - 2)(2x + 3). -(6x^2 + 5x - 6); pair 9 and -4: 3x(2x + 3) - 2(2x + 3).
- (x + 2y)(4x + 3y). ac = 24, pair 3 and 8: x(4x + 3y) + 2y(4x + 3y).
- (5a + 3b)(2a - 3b). ac = -90, pair -15 and 6: 5a(2a - 3b) + 3b(2a - 3b).
- Does not factor. ac = 8, both positive: 1 + 8 = 9 and 2 + 4 = 6, never 5.
- Does not factor. ac = -12: the opposite-sign pairs add to \pm 11, \pm 4 or \pm 1, never -7.
- (2x + 1)(3x - 4). ac = -24, pair -8 and 3: 2x(3x - 4) + 1(3x - 4).
- b = 7, 5, -5, -7. ac = 6, and the pairs 1, 6 and 2, 3 and their negatives add to these.
- If 2x + 6 = 2(x + 3) were a factor, the trinomial would be 2(x + 3) times another binomial, so 2 would divide every one of its terms, a common factor after all.