In many ways, factoring is about patterns—if you acknowledge the patterns that number make once they are multiplied together, you deserve to use those trends to different these numbers right into their individual factors.

You are watching: How to factor x^3 + 1

Some interesting patterns arise once you space working v cubed quantities within polynomials. Special, there room two more special cases to consider: a3 + b3 and also a3 – b3.

Let’s take a watch at exactly how to factor sums and also differences that cubes.

The hatchet “cubed” is used to explain a number elevated to the 3rd power. In geometry, a cube is a six-sided shape with equal width, length, and also height; because all these procedures are equal, the volume of a cube with width x deserve to be stood for by x3. (Notice the exponent!)

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

Before looking in ~ factoring a amount of 2 cubes, stop look at the possible factors.

It turns out the a3 + b3 have the right to actually it is in factored together (a + b)(a2 – abdominal + b2). Let’s examine these components by multiplying.

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

Did you watch that? four of the state cancelled out, leaving us v the (seemingly) basic binomial a3 + b3. So, the determinants are correct.

You have the right to use this sample to variable binomials in the form a3 + b3, otherwise well-known as “the sum of cubes.”

 The amount of Cubes A binomial in the form a3 + b3 have the right to be factored together (a + b)(a2 – abdominal muscle + b2). Examples: The factored form 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 the this binomial fits the amount of cubes pattern: a3 + b3. a = x, and b = 2y (since 2y • 2y • 2y = 8y3). (x + 2y)(x2 – x(2y) + (2y)2) Factor the binomial as (a + b)(a2 – abdominal muscle + b2), substituting a = x and b = 2y right 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 that’s it. The binomial x3 + 8y3 can be factored as (x + 2y)(x2 – 2xy + 4y2)! Let’s shot another one.

You should constantly look because that a typical factor before you follow any kind of of the trends for factoring.

 Example Problem Factor 16m3 + 54n3. 16m3 + 54n3 Factor out the usual factor 2. 2(8m3 + 27n3) 8m3 and also 27n3 room cubes, so you can aspect 8m3 + 27n3 as the sum 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 also b = 3n into the expression (a + b)(a2 – abdominal + b2). 2(2m + 3n)<4m2 – (2m)(3n) + 9n2> Square: (2m)2 = 4m2 and (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. Check your values for a and also 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 4 is the cube root of 64. Substituting these worths for a and also b, you discover (5x + 4)(25x2 – 20x + 16).

C) (x + 4)(x2 – 2x + 16)

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

D) (5x + 4)(25x2 + 20x – 64)

Incorrect. Check the mathematics signs; the b2 ax is positive, no negative, when factoring a amount of cubes. The exactly answer is (5x + 4)(25x2 – 20x + 16).

Difference the Cubes

Having seen exactly how binomials in the kind a3 + b3 deserve to be factored, it need to not come as a surprised that binomials in the type a3 – b3 deserve to be factored in a similar way.

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

Notice the the straightforward construction that the administrate is the exact same as the is because that the sum of cubes; the distinction is in the + and also – signs. Take a minute to compare the factored form of a3 + b3 v the factored type of a3 – b3.

 Factored kind of a3 + b3: (a + b)(a2 – abdominal muscle + b2) Factored form of a3 – b3: (a – b)(a2 + abdominal muscle + b2)

This have the right to be tricky come remember because of the different signs—the factored form of a3 + b3 has a negative, and the factored kind of a3 – b3 consists of a positive! Some civilization remember the different forms favor this:

“Remember one sequence of variables: a3 b3 = (a b)(a2 abdominal muscle b2). There room 4 missing signs. Every little thing the an initial sign is, the is likewise the second sign. The third sign is the opposite, and the fourth sign is constantly +.”

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

Let’s walk ahead and also look at a pair of examples. Mental to element out all common factors first.

 Example Problem Factor 8x3 – 1,000. 8(x3 – 125) Factor out 8. 8(x3 – 125) Identify that the binomial fits the sample a3 - b3: a = x, and also b = 5 (since 53 = 125). 8(x - 5) Factor x3 – 125 as (a – b)(a2 + abdominal muscle + 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) as 5x. Answer 8(x – 5)(x2 + 5x + 25)

Let’s watch what wake up if friend don’t element out the common factor first. In this example, it have the right to still be factored together the difference of two cubes. However, the factored type still has common factors, which should be factored out.

 Example Problem Factor 8x3 – 1,000. 8x3 – 1,000 Identify the 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 + ab + b2), substituting a = 2x and also b = 10 into the expression. (2x – 10)(4x2 + 20x + 100) Square and also multiply: (2x)2 = 4x2, (2x)(10) = 20x, and also 102 = 100. 2(x – 5)(4)(x2 + 5x + 25) Factor the end remaining typical factors in each factor. Factor out 2 indigenous the an initial factor, element out 4 indigenous the second factor. (2 • 4)(x – 5)(x2 + 5x + 25) Multiply the number factors. Answer 8(x – 5)(x2 + 5x + 25)

As you have the right to see, this last example still worked, but required a pair of extra steps. That is always a great idea to element out all usual factors first. In part cases, the just efficient method to element the binomial is to aspect out the typical factors first.

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

 Example Problem Factor r9 – 8s6. r9 – 8s6 Identify this binomial as the difference of two cubes. As shown above, the 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 together (2s2)3. Now the binomial is written in regards to cubed quantities. Thinking of a3 – b3, a = r3 and also 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)