In many ways, factoring is around patterns—if you acknowledge the patterns that number make when they are multiplied together, you have the right to use those patterns to different these numbers right into their separation, personal, instance factors.

You are watching: Factor x^3 + 64

Some amazing patterns arise as soon as you space working through cubed amounts within polynomials. Specifics there room two more special situations to consider: a3 + b3 and also a3 – b3.

Let’s take a look at at how to variable sums and differences the cubes.

The term “cubed” is used to describe a number elevated to the 3rd power. In geometry, a cube is a six-sided shape with equal width, length, and height; due to the fact that all these measures are equal, the volume of a cube v width x deserve to be represented by x3. (Notice the exponent!)

Cubed numbers get huge very quickly. 13 = 1, 23 = 8, 33 = 27, 43 = 64, and also 53 = 125.

Before looking at factoring a sum of 2 cubes, stop look at the feasible factors.

It transforms out that a3 + b3 deserve to actually it is in factored as (a + b)(a2 – abdominal + b2). Let’s check these components by multiplying.

 walk (a + b)(a2 – ab + b2) = a3 + b3? (a)(a2 – abdominal + b2) + (b)(a2 – ab +b2) Apply the distributive property. (a3 – a2b + ab2) + (b)(a2 - ab + b2) Multiply by a. (a3 – a2b + ab2) + (a2b – ab2 + b3) Multiply through b. a3 – a2b + a2b + ab2 – ab2 + b3 Rearrange terms in stimulate to incorporate the prefer terms. a3 + b3 Simplify

Did you see that? 4 of the state cancelled out, leaving us v the (seemingly) an easy binomial a3 + b3. So, the factors are correct.

You can use this sample to aspect binomials in the form a3 + b3, otherwise recognized as “the sum of cubes.”

 The amount of Cubes A binomial in the kind a3 + b3 deserve to be factored together (a + b)(a2 – abdominal muscle + b2). Examples: The factored type of x3 + 64 is (x + 4)(x2 – 4x + 16). The factored kind of 8x3 + y3 is (2x + y)(4x2 – 2xy + y2).

 Example Problem Factor x3 + 8y3. x3 + 8y3 Identify that this binomial fits the sum of cubes pattern: a3 + b3. a = x, and also b = 2y (since 2y • 2y • 2y = 8y3). (x + 2y)(x2 – x(2y) + (2y)2) Factor the binomial as (a + b)(a2 – abdominal + b2), substituting a = x and also b = 2y into the expression. (x + 2y)(x2 – x(2y) + 4y2) Square (2y)2 = 4y2. Answer (x + 2y)(x2 – 2xy + 4y2) Multiply −x(2y) = −2xy (writing the coefficient first.

And it is it. The binomial x3 + 8y3 have the right to be factored together (x + 2y)(x2 – 2xy + 4y2)! Let’s try another one.

You should always look for a usual factor prior to you follow any type of of the trends for factoring.

 Example Problem Factor 16m3 + 54n3. 16m3 + 54n3 Factor the end the typical factor 2. 2(8m3 + 27n3) 8m3 and also 27n3 are cubes, so friend can aspect 8m3 + 27n3 together the amount of 2 cubes: a = 2m, and also b = 3n. 2(2m + 3n)<(2m)2 – (2m)(3n) + (3n)2> Factor the binomial 8m3 + 27n3 substituting a = 2m and b = 3n right into the expression (a + b)(a2 – abdominal + b2). 2(2m + 3n)<4m2 – (2m)(3n) + 9n2> Square: (2m)2 = 4m2 and also (3n)2 = 9n2. Answer 2(2m + 3n)(4m2 – 6mn + 9n2) Multiply −(2m)(3n) = −6mn.

Factor 125x3 + 64.

A) (5x + 64)(25x2 – 125x + 16)

B) (5x + 4)(25x2 – 20x + 16)

C) (x + 4)(x2 – 2x + 16)

D) (5x + 4)(25x2 + 20x – 64)

A) (5x + 64)(25x2 – 125x + 16)

Incorrect. Inspect your worths for a and b here. B3 = 64, so what is b? The correct answer is (5x + 4)(25x2 – 20x + 16).

B) (5x + 4)(25x2 – 20x + 16)

Correct. 5x is the cube root of 125x3, and also 4 is the cube source of 64. Substituting these worths for a and b, you find (5x + 4)(25x2 – 20x + 16).

C) (x + 4)(x2 – 2x + 16)

Incorrect. Inspect your worths for a and also b here. A3 = 125x3, so what is a? The exactly answer is (5x + 4)(25x2 – 20x + 16).

D) (5x + 4)(25x2 + 20x – 64)

Incorrect. Check the math signs; the b2 ax is positive, not negative, once factoring a amount of cubes. The exactly answer is (5x + 4)(25x2 – 20x + 16).

Difference the Cubes

Having seen just how binomials in the kind a3 + b3 deserve to be factored, it should not come as a surprised that binomials in the form a3 – b3 have the right to be factored in a comparable way.

 The difference of Cubes A binomial in the kind a3 – b3 deserve to be factored together (a – b)(a2 + abdominal muscle + b2). Examples: The factored kind of x3 – 64 is (x – 4)(x2 + 4x + 16). The factored type of 27x3 – 8y3 is (3x – 2y)(9x2 + 6xy + 4y2).

Notice that the simple construction that the factorization is the very same as it is for the amount of cubes; the distinction is in the + and – signs. Take it a minute to compare the factored kind of a3 + b3 through the factored form of a3 – b3.

 Factored type of a3 + b3: (a + b)(a2 – ab + b2) Factored kind of a3 – b3: (a – b)(a2 + ab + b2)

This can be tricky come remember due to the fact that of the various signs—the factored kind of a3 + b3 has a negative, and the factored form of a3 – b3 has a positive! Some people remember the different forms like this:

“Remember one sequence of variables: a3 b3 = (a b)(a2 abdominal b2). There space 4 lacking signs. Every little thing the first sign is, that is also the second sign. The 3rd sign is the opposite, and the fourth sign is always +.”

Try this because that yourself. If the first sign is +, together in a3 + b3, according to this strategy exactly how do you to fill in the rest: (a b)(a2 abdominal b2)? does this technique help girlfriend remember the factored kind of a3 + b3 and a3 – b3?

Let’s go ahead and look in ~ a pair of examples. Mental to variable out all usual factors first.

 Example Problem Factor 8x3 – 1,000. 8(x3 – 125) Factor the end 8. 8(x3 – 125) Identify that the binomial fits the pattern a3 - b3: a = x, and also b = 5 (since 53 = 125). 8(x - 5) Factor x3 – 125 as (a – b)(a2 + abdominal + b2), substituting a = x and also b = 5 into the expression. 8(x – 5)(x2 + 5x + 25) Square the first and last terms, and also rewrite (x)(5) together 5x. Answer 8(x – 5)(x2 + 5x + 25)

Let’s watch what wake up if friend don’t element out the usual factor first. In this example, it can still be factored together the distinction of 2 cubes. However, the factored kind still has usual factors, which have to be factored out.

 Example Problem Factor 8x3 – 1,000. 8x3 – 1,000 Identify that this binomial fits the pattern a3 - b3: a = 2x, and also b = 10 (since 103 = 1,000). (2x – 10)<(2x)2 + 2x(10) + 102> Factor together (a – b)(a2 + abdominal muscle + b2), substituting a = 2x and also b = 10 into the expression. (2x – 10)(4x2 + 20x + 100) Square and multiply: (2x)2 = 4x2, (2x)(10) = 20x, and 102 = 100. 2(x – 5)(4)(x2 + 5x + 25) Factor the end remaining typical factors in each factor. Element out 2 from the an initial factor, variable out 4 indigenous the 2nd factor. (2 • 4)(x – 5)(x2 + 5x + 25) Multiply the numerical factors. Answer 8(x – 5)(x2 + 5x + 25)

As you can see, this last instance still worked, but required a pair of extra steps. That is always a great idea to aspect out all common factors first. In part cases, the only efficient method to factor the binomial is to aspect out the usual factors first.

Here is one an ext example. Note that r9 = (r3)3 and also that 8s6 = (2s2)3.

 Example Problem Factor r9 – 8s6. r9 – 8s6 Identify this binomial together the difference of 2 cubes. As displayed above, that is. Making use of the laws of exponents, rewrite r9 together (r3)3. (r3)3 – (2s2)3 Rewrite r9 together (r3)3 and also rewrite 8s6 as (2s2)3. Now the binomial is written in regards to cubed quantities. Reasoning of a3 – b3, a = r3 and b = 2s2. (r3 – 2s2)<(r3)2 + (r3)(2s2) + (2s2)2> Factor the binomial as  (a – b)(a2 + ab + b2), substituting a = r3 and also b = 2s2 into the expression. (r3 – 2s2)(r6 + 2 r3s2+ 4s4) Multiply and also square the terms. Answer (r3 – 2s2)(r6 + 2r3s2 + 4s4)