How do you factor #x^3+x^2-x-1#?
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We must first determine what numbers are factors using the rational root theorem and the remainder theorem.
The rational root theorem helps us list all possible factors of a particular polynomial of the form #f(x) = qx^n + mx^(n - 1) + ... + p#. The roots are given by #("factors of p")/("factors of q")#.
In our case, the factors of #-1# are #+-1# and the factors of #1# are #+-1#. Hence, the rational root theorem says that our possible factors are #(+-1)/(+-1) = +-1#.
The remainder theorem helps to see if a given number is a factor without having to effectuate the full division. If #x - a# is a factor of #f(x) = qx^n + mx^(n - 1) + ... + p#, then #f(a) = 0#.
Let's start with the positive #1#.
#1^3 + 1^2 - 1 - 1 =^? 0#
#2 - 2 = 0#
Yes, #x - 1# is a factor of #x^3 + x^2 - x - 1#!
Now, we can use either synthetic or long division to start the factoring process. I will use synthetic, because for me it is extremely efficient and quick at getting the job done.
#"1_|1 1 -1 -1"#
#" 1 2 1"#
"________"
#" 1 2 1 0"#
Hence, #(x^3 + x^2 - x - 1) ÷ (x - 1) = x^2 + 2x + 1#. Otherwise put #(x - 1)(x^2 + 2x + 1) = x^3 + x^2 - x - 1#. We're not done yet, though. #x^2 + 2x + 1# can be factored as a perfect square trinomial, #(a + b)(a + b) = (a + b)^2#.
#x^2 + 2x + 1 = (x + 1)^2#.
So, #x^2 + x^2 - x - 1 = (x - 1)(x + 1)(x + 1) = (x - 1)(x + 1)^2#.
Hopefully this helps!
