Factoring ax^2 + bx + c

Factoring the quadratic expression ax² + bx + c is a common component of most high school math curricula.  The simplest type of factoring problems is when a = 1, for example, x² + 8x + 15.  The algorithm for finding the factored form of x² + bx + c is to find two numbers, p and q, such that pq = c and p + q = b, and writing the solution as (x + p)(x + q).  So for x² + 8x + 15, the two numbers p and q such that pq = 15 and p + q = 8 are 3 and 5, so the solution is (x + 3)(x + 5).  With practice, factoring x² + bx + c is straightforward and can be done mentally.

 The harder factoring problems are when a > 1, for example, 8x² + 14x + 3.  To factor ax² + bx + c when a > 1, the textbook I teach out of (McDougal Littell) says to find factors of a and factors of c, and guess and check with those numbers until it multiplies out to the original question.  So for 8x² + 14x + 3, the factors of 8 are 1 & 8 and 2 & 4, and factors of 3 are 1 & 3.  The different possible combinations (and their solutions) for all these numbers are:

(x + 1)(8x + 3)	= 8x2 + 3x + 8x + 3	= 8x2 + 11x + 3
(x + 3)(8x + 1)	= 8x2 + x + 24x + 3	= 8x2 + 25x + 3
(2x + 1)(4x + 3)	= 8x2 + 6x + 4x + 3	= 8x2 + 10x + 3
(2x + 3)(4x + 1)	= 8x2 + 2x + 12x + 3	= 8x2 + 14x + 3

The only possibility that multiplied with the correct middle term of 14x is the last one, therefore, 8x² + 14x + 3 = (2x + 3)(4x + 1).

For the most part I like the McDougal Littell textbook, and to be fair this is the common algorithm for factoring ax² + bx + c when a > 1, but in my opinion this is not the best way.  It is long and cumbersome, and gets even worse for larger numbers with more factors.  It also relies on trial and error and a little bit of luck, all things good mathematicians should avoid when solving problems.  I can think of several better ways to factor ax² + bx + c when a > 1, and in fact I teach one of the ways to my own high school students.

Method 1 – Factoring b and ac

The first alternate algorithm for factoring ax² + bx + c when a > 1 is the way I learned when I was a student and the way I teach my own high school math students.  First, find two numbers p and q such that pq = ac and p + q = b, and then split bx into two terms px + qx.  Then factor out the first two terms (ax² + px) and then factor out the last two terms (qx + c).  If done correctly, there will be a common term (in parenthesis) which can be combined to give a factored solution to the problem.  For example, to factor 8x² + 14x + 3, first find two numbers p and q such that pq = 8∙3 = 24 and p + q = 14, which are 2 and 12, and then split 14x into 2x and 12x, giving 8x² + 2x + 12x + 3.  Then factor out the first two terms 8x² + 2x to 2x(4x + 1) and the last two terms 12x + 3 to 3(4x + 1), so that 8x² + 2x + 12x + 3 = 2x(4x + 1) + 3(4x + 1).  There is a common term (4x + 1) that can make 2x and 3 combine, so 8x² + 14x + 3 = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8x2 + 2x + 12x + 3	(2 and 12 add up to 14 and multiply to 8∙3)
	= 2x(4x + 1) + 3(4x + 1)	(factor the first two and the last two terms)
	= (2x + 3)(4x + 1)	(combine the like term of 4x + 1)

Reasoning: It is helpful to think of the factored answer as (ex + f)(gx + h), where e, f, g, and h are unknowns that need to be solved.  Multiplying (ex + f)(gx + h) gives egx² + (eh + fg)x + fh, so matching that with ax² + bx + c, a = eg, b = eh + fg, and c = fh.  That means finding two numbers that multiply to ac and add up to b is equivalent to finding eh and fg.  So ax² + bx + c = egx² + ehx + fgx + fh = ex(gx + h) + f(gx + h) = (ex + f)(gx + h).

Method 2 – Factoring with Fractions

The second alternate algorithm for factoring ax² + bx + c when a > 1 is to factor the a out first, making the middle term ^b/_a and the last term ^c/_a, then re-write the last term ^c/_a as ^ac/_a^2, which makes ax² + bx + c = a(x² + ^b/_ax + ^ac/_a^2).  To factor, you need to find two numbers p and q such that pq = ^ac/_a^2 and p + q = ^b/_a, but both p and q will have a denominator of a so really the only mental math to be done is to find the numerators, or in other words, two numbers that multiply to ac and add up to b.  Then re-write ax² + bx + c as a(x + p)(x + q), simplify the fractions, and distribute the factors of a in such a way to remove the fractions.  For example, to factor 8x² + 14x + 3, factor out the 8 first so it is 8(x² + ¹⁴/₈x + ³/₈), and then re-write the last term as ^ac/_a^2 which is ²⁴/₆₄.  Next find two numbers that multiply to 24 and add up to 14, which are 12 and 2, so p = ¹²/_a = ¹²/₈ and q = ²/_a = ²/₈, so 8x² + 14x + 3 = 8(x + ¹²/₈)(x + ²/₈) or simplifying, 8x² + 14x + 3 = 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x2 + 14/8x + 3/8)	(factor out an 8)
	= 8(x2 + 14/8x + 24/64)	(change 3/8 to ac/a^2 = 24/64)
	= 8(x + 12/8)(x + 2/8)	(12 and 2 add up to 14 and multiply to 24)
	= 8(x + 3/2)(x + 1/4)	(12/8 simplified is 3/2, 2/8 simplified is 1/4)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: Factoring out the a term leaves a quadratic expression with a = 1, however with fractions.  Forcing the middle term to have a denominator of a and the last term to have a denominator of a² forces both p and q to have a denominator of a, which makes it easier to find p and q even if they are fractions.

Method 3 – Factoring by Completing the Square

The third way to factor ax² + bx + c when a > 1 is to complete the square.  First factor the a, making ax² + bx + c = a(x² + ^b/_ax + ^c/_a).  Then complete the square on x² + ^b/_ax by adding ^{b^2}/_4a^2 (and subtracting ^{b^2}/_4a^2 to balance the equation), which makes a(x² + ^b/_ax + ^{b^2}/_4a^2 – ^{b^2}/_4a^2 + ^c/_a) = a((x + ^b/_2a)² – (^{b^2}/_4a^2 – ^c/_a)).  If ax² + bx + c is factorable, then ^{b^2}/_4a^2 – ^c/_a should be a square number, so replace it with t², and replace ^b/_2a with s, which makes a((x + s)² – t²).  Factoring this difference of squares makes a(x + s + t)(x + s – t).  Then the factors of the a in front can be distributed in a way to remove the fractions and get the factored solution.  For example, to factor 8x² + 14x + 3, factor out the 8 first so it is 8(x² + ⁷/₄x + ³/₈).  Then complete the square by adding (and subtracting) the term ⁴⁹/₆₄ to make 8(x² + ⁷/₄x + ⁴⁹/₆₄ – ⁴⁹/₆₄ + ³/₈) which is 8((x + ⁷/₈)² – ²⁵/₆₄).  Factoring the difference of squares makes 8(x + ⁷/₈ + ⁵/₈)(x + ⁷/₈ – ⁵/₈) or 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x2 + 7/4x + 3/8)	(factor out an 8)
	= 8(x2 + 7/4x + 49/64 – 49/64 + 3/8)	(complete the square with 49/64)
	= 8((x + 7/8)2 – 25/64)	(finish completing the square)
	= 8(x + 7/8 + 5/8)(x + 7/8 – 5/8)	(factor the difference of squares)
	= 8(x + 3/2)(x + 1/4)	(simplify the fractions)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: ^{b^2}/_4a^2 – ^c/_a is a square number because it is equal to ^{(b^2 – 4ac)}/_4a^2 and in terms of e, f, g, and h (defined in Method 1) it is equal to ^{((eh + fg)^2 – 4egfh)}/_4(eg)^2 = ^{(e^2h^2 + 2efgh + f^2g^2 – 4efgh)}/_4e^2g^2 = ^{(e^2h^2 – 2efgh + f^2g^2)}/_4e^2g^2 = ^{(eh – fg)^2}/_(2eg)^2.

Method 4 – Factoring with the Quadratic Equation

The fourth way to factor ax² + bx + c when a > 1 is to use the quadratic formula, x = ^{(–b ± √(b^2 – 4ac))}/_2a, to show that ax² + bx + c = a(x – ^{(–b – √(b^2 – 4ac))}/_2a)(x – ^{(–b + √(b^2 – 4ac))}/_2a).  Then the factors of the a in front can be distributed in a way to remove the fractions and get the factored solution.  For example, 8x² + 14x + 3 = a(x – ^{(–b – √(b^2 – 4ac))}/_2a)(x – ^{(–b + √(b^2 – 4ac))}/_2a) = 8(x – ^{(–14 – √(14^2 – 4∙8∙3))}/_2∙8)(x – ^{(–14 + √(14^2 – 4∙8∙3))}/_2∙8) = 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x – (–b – √(b^2 – 4ac))/2a)	(formula)
	∙(x – (–b + √(b^2 – 4ac))/2a)
	= 8(x – (–14 – √(14^2 – 4∙8∙3))/2∙8)	(plug in a = 8, b = 14, c = 3)
	∙(x – (–14 + √(14^2 – 4∙8∙3))/2∙8)
	= 8(x + 3/2)(x + 1/4)	(solve)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: The quadratic equation can be proved by completing the square, and so can this method.  In Method 3 we already showed that ax² + bx + c = a((x + ^b/_2a)² – (^{b^2}/_4a^2 – ^c/_a)).  Continuing on, ^{b^2}/_4a^2 – ^c/_a = ^{(b^2 – 4ac)}/_4a^2 = (^{√(b^2 – 4ac)}/_2a)² so a((x + ^b/_2a)² – (^{b^2}/_4a^2 – ^c/_a)) = a((x + ^b/_2a)² – (^{√(b^2 – 4ac)}/_2a)²).  Factoring the difference of squares gives a(x + ^{(b + √(b^2 – 4ac))}/_2a)(x + ^{(b – √(b^2 – 4ac))}/_2a) = a(x – ^{(–b – √(b^2 – 4ac))}/_2a)(x – ^{(–b + √(b^2 – 4ac))}/_2a).

Method 5 – Factoring by Graphing

The fifth way to factor ax² + bx + c when a > 1 is to graph the equation and use the x-intercepts.  Unfortunately this method relies on a graphing calculator, but it is fairly quick if a graphing calculator is available.  In this method, graph the function y = ax² + bx + c and find p and q, the two x-intercepts (as fractions).  Then ax² + bx + c can be re-written as a(x – p)(x – q), and the factors of a can be distributed in a way to remove the fractions p and q and get the factored solution.  So for 8x² + 14x + 3, you would graph and find the two x-intercepts, which are x = -1.5 = -³/₂ and x = -0.25 = -¹/₄. 

This means 8x² + 14x + 3 = 8(x – -³/₂)(x – -¹/₄) = 8(x + ³/₂)(x + ¹/₄).  The 8 in front can be factored as 2 and 4 to get rid of both denominators in the parentheses, and distributing 2(x + ³/₂)4(x + ¹/₄) = (2x + 3)(4x + 1).

8x2 + 14x + 3	= 8(x – -1.5)(x – -0.25)	(two x-intercepts)
	= 8(x + 3/2)(x + 1/4)	(re-write as fractions)
	= 2(x + 3/2)4(x + 1/4)	(3/2 had a den of 2, 1/4 has a den of 4)
	= (2x + 3)(4x + 1)	(distribute the 2 and 4)

Reasoning: Since the x-intercepts = ^{(–b ± √(b^2 – 4ac))}/_2a, this method works for the same reason using the quadratic equation works, which is described in Method 4.

Conclusion

As you can see, there are several different ways to factor ax² + bx + c where a > 1: factoring b and ac, factoring with fractions, factoring by completing the square, factoring using the quadratic equation, and factoring by graphing.  (There are probably other methods as well that I have not considered.)  I feel that all these alternate methods are superior to the method of guessing and checking, which does not seem very mathematical at all.