If you mean 1/2 * (x-2)^3 + 1 then we have that pesky 1/2 to deal with.
(a^3 + b^3) = (a+b)(a^2 - ab + b^2)
So, let a = cbrt(1/2)*x and b = 1, or 2^(1/3)*x
Then 1/2 * (x-2)^3 + 1
= (x*2^1/3 + 1)(x^2 * 2^2/3 - x*2^1/3 + 1)
or,
(cbrt(2)x + 1)(cbrt(4)x^2 - cbrt(2)x + 1)
How do you factor this: f(x)=1/2(x-2)^3+1
5 answers
Oops. Those 2's in the solution should all be 1/2.
Can you explain it more clearly?
What does cbrt mean?
sqrt = square root
cbrt = cube root.
The fractional exponents get clumsy
suppose you had
125x^3 + 1
that is (5x)^3 + 1^3
so it factors into
(5x+1)((5x)^2 - (5x)*1 + 1^2)
= (5x+1)(25x^2 - 5x + 1)
You have (1/2)x^3 which is ((1/2)^(1/3)x)^3 or (cbrt(1/2)x)^3
cbrt = cube root.
The fractional exponents get clumsy
suppose you had
125x^3 + 1
that is (5x)^3 + 1^3
so it factors into
(5x+1)((5x)^2 - (5x)*1 + 1^2)
= (5x+1)(25x^2 - 5x + 1)
You have (1/2)x^3 which is ((1/2)^(1/3)x)^3 or (cbrt(1/2)x)^3