Completely factorize 8 - (3 - x)^3 ?

You factor this by the difference of two cubes: a^3 - b^3 = (a - b)(a^2 + ab + b^2). For this, the cube root of 8 is 2, so let that be a. The cube root of (3 - x)^3 is (3 - x), so let that be b.

First start off with the (a - b) part. You would have (2 - (3 - x)), which is then (2 - 3 + x) = (-1 + x) or (x - 1).

Now for the (a^2 + ab + b^2). a^2 would be 2^2 which equals 4. ab would be 2(3 - x) = 6 - 2x. And b^2 would be (3 - x)^2 = 9 - 6x + x^2 or x^2 - 6x + 9. So altogether, a^2 + ab + b^2 is 4 + 6 - 2x + x^2 - 6x + 9 = x^2 - 8x + 19.

And so 8 - (3 - x)^3 factored is (x - 1)(x^2 - 8x + 19). Whenever you factor something by the difference of two cubes, nothing is further able to be factored, so don't even bother trying to factor x^2 - 8x + 19 ;).

ANSWER: (x - 1)(x^2 - 8x + 19)

8 - (3 - x)^3

= 2^3 - (3 - x)^3

This is a difference of cubes:

= (2 - (3 - x))(4 + 2(3 - x) + (3 - x)^2)

= (2 - 3 + x) ( 4 + 6 - 2x + 9 - 6x + x^2)

= (x - 1)(19 - 8x + x^2)

= (x - 1)(x^2 - 8x + 19)

a) locate p and q such that p+q = 18 and pq = 9(-sixteen) = -a hundred and forty four we get p = 24 and q = -6 so, 9t² + 18t -sixteen = (9t +24)(9t - 6)/9 = 3(3t + 8)3(3t - 2)/9 = (3t + 8)(3t - 2) b) undergo in ideas that A² - B² = (A + B)(A - B) 16x^4 - 81y^8 = (4x²)² - (9y^4)² = (4x² + 9y²) (4x² - 9y^4) = (4x² + 9y²) (2x + 3y²)(2x - 3y²) c) x³ + x² - 6x = x(x² + x - 6) returned, like variety a), locate p and q such that p + q = a million and pq = -6 we get p = 3 and q = -2 so = x (x + 3)(x - 2) carried out.