Expres­sions such as x3+8x^3 + 8, x4−1x^4 - 1 and x5−32x^5 - 32 appear again and again, and so do num­ber ques­tions like "find the remain­der when 13141+1114113^{141} + 11^{141} is divided by 2424". All of them are about one fam­ily, xn+anx^n + a^n and xn−anx^n - a^n. The fac­tor the­o­rem tells us exactly when x+ax + a and x−ax - a divide these expres­sions, and a pat­tern gives the other fac­tor. This les­son col­lects those results, which you will use for the rest of your alge­bra.

Result 1: x + a divides xⁿ + aⁿ when n is odd

x+ax + a is a fac­tor of xn+anx^n + a^n for every odd pos­i­tive inte­ger nn.

Proof. Let f(x)=xn+anf(x) = x^n + a^n. By the fac­tor the­o­rem we test x=−ax = -a:

f(−a)=(−a)n+an=−an+an=0(since n is odd)f(-a) = (-a)^n + a^n = -a^n + a^n = 0 \quad \text{(since } n \text{ is odd)}

So x+ax + a is a fac­tor.

The other fac­tor fol­lows a pat­tern of alter­nat­ing signs, with the power of xx falling and the power of aa ris­ing:

xn+an=(x+a)(xn−1−xn−2a+xn−3a2−⋯+an−1),n oddx^n + a^n = (x + a)\left(x^{n-1} - x^{n-2}a + x^{n-3}a^2 - \dots + a^{n-1}\right), \quad n \text{ odd}

For exam­ple,

x3+a3=(x+a)(x2−xa+a2)x^3 + a^3 = (x + a)(x^2 - xa + a^2)

x5+y5=(x+y)(x4−x3y+x2y2−xy3+y4)x^5 + y^5 = (x + y)(x^4 - x^3y + x^2y^2 - xy^3 + y^4)

The first of these is the famil­iar iden­tity a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab + b^2).

What if nn is even? Then f(−a)=(−a)n+an=an+an=2anf(-a) = (-a)^n + a^n = a^n + a^n = 2a^n, which is not zero (for a≠0a \neq 0). So x+ax + a is not a fac­tor of xn+anx^n + a^n when nn is even. For instance, x2+1x^2 + 1 has no fac­tor x+1x + 1.

Result 2: x − a divides xⁿ − aⁿ for every n

x−ax - a is a fac­tor of xn−anx^n - a^n for every pos­i­tive inte­ger nn.

Proof. Let f(x)=xn−anf(x) = x^n - a^n. Then f(a)=an−an=0f(a) = a^n - a^n = 0, so x−ax - a is a fac­tor.

This time all the signs in the other fac­tor are plus:

xn−an=(x−a)(xn−1+xn−2a+xn−3a2+⋯+an−1)x^n - a^n = (x - a)\left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + a^{n-1}\right)

For exam­ple,

x2−a2=(x−a)(x+a)x^2 - a^2 = (x - a)(x + a)

x3−a3=(x−a)(x2+xa+a2)x^3 - a^3 = (x - a)(x^2 + xa + a^2)

x4−a4=(x−a)(x3+x2a+xa2+a3)x^4 - a^4 = (x - a)(x^3 + x^2a + xa^2 + a^3)

Result 3: both x + a and x − a divide xⁿ − aⁿ when n is even

x+ax + a and x−ax - a are both fac­tors of xn−anx^n - a^n for every even pos­i­tive inte­ger nn.

We already know x−ax - a is a fac­tor. For x+ax + a, test x=−ax = -a: f(−a)=(−a)n−an=an−an=0f(-a) = (-a)^n - a^n = a^n - a^n = 0, because an even power of −a-a is pos­i­tive.

x2−a2=(x+a)(x−a)x^2 - a^2 = (x + a)(x - a)

x4−a4=(x2+a2)(x2−a2)=(x2+a2)(x+a)(x−a)x^4 - a^4 = (x^2 + a^2)(x^2 - a^2) = (x^2 + a^2)(x + a)(x - a)

Worked exam­ples

Exam­ple 1

Fac­torise x4−1x^4 - 1 in two ways.

Using Result 2 with a=1a = 1:

x4−1=x4−14=(x−1)(x3+x2+x+1)x^4 - 1 = x^4 - 1^4 = (x - 1)(x^3 + x^2 + x + 1)

Using the dif­fer­ence of squares twice:

x4−1=(x2)2−12=(x2+1)(x2−1)=(x2+1)(x+1)(x−1)x^4 - 1 = (x^2)^2 - 1^2 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1)

The sec­ond form is com­plete. Com­par­ing the two shows that x3+x2+x+1=(x+1)(x2+1)x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1).

Exam­ple 2

Fac­torise x5−1x^5 - 1 and x5+1x^5 + 1.

x5−1=x5−15=(x−1)(x4+x3+x2+x+1)x^5 - 1 = x^5 - 1^5 = (x - 1)(x^4 + x^3 + x^2 + x + 1)

Since 55 is odd, Result 1 applies to x5+1x^5 + 1:

x5+1=(x+1)(x4−x3+x2−x+1)x^5 + 1 = (x + 1)(x^4 - x^3 + x^2 - x + 1)

Exam­ple 3

Fac­torise x3+1x^3 + 1 and x3+8x^3 + 8.

x3+1=x3+13=(x+1)(x2−x+1)x^3 + 1 = x^3 + 1^3 = (x + 1)(x^2 - x + 1)

x3+8=x3+23=(x+2)(x2−2x+4)x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)

In the same way x+2x + 2 is a fac­tor of x5+32x^5 + 32, because 32=2532 = 2^5 and 55 is odd.

Remain­ders of large num­bers

The results hold for num­bers too, because sub­sti­tut­ing whole num­bers for xx and aa keeps every coef­fi­cient of the other fac­tor a whole num­ber.

Exam­ple 4

Find the remain­der when 13141+1114113^{141} + 11^{141} is divided by 2424.

Take x=13x = 13, a=11a = 11 and n=141n = 141, which is odd. By Result 1, x+a=13+11=24x + a = 13 + 11 = 24 is a fac­tor of 13141+1114113^{141} + 11^{141}. So the remain­der is 00.

Exam­ple 5

Find the remain­der when x11−1x^{11} - 1 is divided by x−1x - 1.

By Result 2, x−1x - 1 is a fac­tor of x11−1x^{11} - 1, so the remain­der is 00. (Directly: f(1)=1−1=0f(1) = 1 - 1 = 0.)

Prac­tice

  1. Fac­torise x3+27x^3 + 27.
  2. Fac­torise x5−32x^5 - 32.
  3. Fac­torise x4−81x^4 - 81 com­pletely.
  4. Fac­torise x5+243x^5 + 243.
  5. Fac­torise x6−1x^6 - 1 com­pletely.
  6. Show that x−1x - 1 is a fac­tor of both x10−1x^{10} - 1 and x11−1x^{11} - 1.
  7. Is x+1x + 1 a fac­tor of x10+1x^{10} + 1? If not, find the remain­der.
  8. Find the remain­der when 725+5257^{25} + 5^{25} is divided by 1212.
  9. Find the remain­der when 1720−132017^{20} - 13^{20} is divided by 3030.
  10. Find the remain­der when x51+51x^{51} + 51 is divided by x+1x + 1.
  11. Find kk if x140+2x151+kx^{140} + 2x^{151} + k is divis­i­ble by x+1x + 1.

Answers

Show answers
  1. x3+33=(x+3)(x2−3x+9)x^3 + 3^3 = (x + 3)(x^2 - 3x + 9)
  2. x5−25=(x−2)(x4+2x3+4x2+8x+16)x^5 - 2^5 = (x - 2)(x^4 + 2x^3 + 4x^2 + 8x + 16)
  3. (x2+9)(x2−9)=(x2+9)(x+3)(x−3)(x^2 + 9)(x^2 - 9) = (x^2 + 9)(x + 3)(x - 3)
  4. x5+35=(x+3)(x4−3x3+9x2−27x+81)x^5 + 3^5 = (x + 3)(x^4 - 3x^3 + 9x^2 - 27x + 81)
  5. (x3−1)(x3+1)=(x−1)(x2+x+1)(x+1)(x2−x+1)(x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)
  6. With f(x)=x10−1f(x) = x^{10} - 1, f(1)=1−1=0f(1) = 1 - 1 = 0; with g(x)=x11−1g(x) = x^{11} - 1, g(1)=1−1=0g(1) = 1 - 1 = 0. (Result 2 for n=10n = 10 and n=11n = 11.)
  7. No. f(−1)=(−1)10+1=2f(-1) = (-1)^{10} + 1 = 2, so the remain­der is 22.
  8. 2525 is odd, so 7+5=127 + 5 = 12 is a fac­tor: remain­der 00.
  9. 2020 is even, so 17+13=3017 + 13 = 30 is a fac­tor of 1720−132017^{20} - 13^{20}: remain­der 00.
  10. f(−1)=(−1)51+51=−1+51=50f(-1) = (-1)^{51} + 51 = -1 + 51 = 50
  11. f(−1)=1−2+k=0f(-1) = 1 - 2 + k = 0, so k=1k = 1.