Expressions such as x3+8, x4−1 and x5−32 appear again and again, and so do number questions like "find the remainder when 13141+11141 is divided by 24". All of them are about one family, xn+an and xn−an. The factor theorem tells us exactly when x+a and x−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+a is a factor of xn+an for every odd positive integer n.
Proof. Let f(x)=xn+an. By the factor theorem we test x=−a:
f(−a)=(−a)n+an=−an+an=0(since n is odd)
So x+a is a factor.
The other factor follows a pattern of alternating signs, with the power of x falling and the power of a rising:
xn+an=(x+a)(xn−1−xn−2a+xn−3a2−⋯+an−1),n odd
For example,
x3+a3=(x+a)(x2−xa+a2)
x5+y5=(x+y)(x4−x3y+x2y2−xy3+y4)
The first of these is the familiar identity a3+b3=(a+b)(a2−ab+b2).
What if n is even? Then f(−a)=(−a)n+an=an+an=2an, which is not zero (for a=0). So x+a is not a factor of xn+an when n is even. For instance, x2+1 has no factor x+1.
Result 2: x − a divides xⁿ − aⁿ for every n#
x−a is a factor of xn−an for every positive integer n.
Proof. Let f(x)=xn−an. Then f(a)=an−an=0, so x−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)
For example,
x2−a2=(x−a)(x+a)
x3−a3=(x−a)(x2+xa+a2)
x4−a4=(x−a)(x3+x2a+xa2+a3)
Result 3: both x + a and x − a divide xⁿ − aⁿ when n is even#
x+a and x−a are both factors of xn−an for every even positive integer n.
We already know x−a is a factor. For x+a, test x=−a: f(−a)=(−a)n−an=an−an=0, because an even power of −a is positive.
x2−a2=(x+a)(x−a)
x4−a4=(x2+a2)(x2−a2)=(x2+a2)(x+a)(x−a)
Worked examples#
Example 1#
Factorise x4−1 in two ways.
Using Result 2 with a=1:
x4−1=x4−14=(x−1)(x3+x2+x+1)
Using the difference of squares twice:
x4−1=(x2)2−12=(x2+1)(x2−1)=(x2+1)(x+1)(x−1)
The second form is complete. Comparing the two shows that x3+x2+x+1=(x+1)(x2+1).
Example 2#
Factorise x5−1 and x5+1.
x5−1=x5−15=(x−1)(x4+x3+x2+x+1)
Since 5 is odd, Result 1 applies to x5+1:
x5+1=(x+1)(x4−x3+x2−x+1)
Example 3#
Factorise x3+1 and x3+8.
x3+1=x3+13=(x+1)(x2−x+1)
x3+8=x3+23=(x+2)(x2−2x+4)
In the same way x+2 is a factor of x5+32, because 32=25 and 5 is odd.
Remainders of large numbers#
The results hold for numbers too, because substituting whole numbers for x and a keeps every coefficient of the other factor a whole number.
Example 4#
Find the remainder when 13141+11141 is divided by 24.
Take x=13, a=11 and n=141, which is odd. By Result 1, x+a=13+11=24 is a factor of 13141+11141. So the remainder is 0.
Example 5#
Find the remainder when x11−1 is divided by x−1.
By Result 2, x−1 is a factor of x11−1, so the remainder is 0. (Directly: f(1)=1−1=0.)
Practice#
- Factorise x3+27.
- Factorise x5−32.
- Factorise x4−81 completely.
- Factorise x5+243.
- Factorise x6−1 completely.
- Show that x−1 is a factor of both x10−1 and x11−1.
- Is x+1 a factor of x10+1? If not, find the remainder.
- Find the remainder when 725+525 is divided by 12.
- Find the remainder when 1720−1320 is divided by 30.
- Find the remainder when x51+51 is divided by x+1.
- Find k if x140+2x151+k is divisible by x+1.
Answers#
Show answers
- x3+33=(x+3)(x2−3x+9)
- x5−25=(x−2)(x4+2x3+4x2+8x+16)
- (x2+9)(x2−9)=(x2+9)(x+3)(x−3)
- x5+35=(x+3)(x4−3x3+9x2−27x+81)
- (x3−1)(x3+1)=(x−1)(x2+x+1)(x+1)(x2−x+1)
- With f(x)=x10−1, f(1)=1−1=0; with g(x)=x11−1, g(1)=1−1=0. (Result 2 for n=10 and n=11.)
- No. f(−1)=(−1)10+1=2, so the remainder is 2.
- 25 is odd, so 7+5=12 is a factor: remainder 0.
- 20 is even, so 17+13=30 is a factor of 1720−1320: remainder 0.
- f(−1)=(−1)51+51=−1+51=50
- f(−1)=1−2+k=0, so k=1.