24x^8-21x^5-3x^2

Question

This question asks to factor the trinomial completely.  Im in pre calculus class in college, help would be greatly appreciated :)

Answer 1

First, factor out the greatest common factor, 3x², to get

 

3x²(8x⁶ -7x³ -1)

 

Notice that x⁶ is the square of x³, so if we let u = x³, the expression becomes 3x²(8u² - 7u - 1)

                           = 3x²(8u + 1)(u - 1)

                           = 3x²(8x³ + 1)(x³ - 1)

 

A³ + B³  = (A + B)(A² - AB + B²)

A³ - B³ = (A - B)(A² + AB + B²)

 

So, 8x³ + 1 = (2x)³ + 1³  = (2x + 1)(4x² - 2x + 1)

 and x³ - 1 = (x - 1)(x² + x + 1)

 

Therefore, 

 

24x⁸ - 21x⁵ - 3x² =

         

                 3x²(2x + 1)(4x² - 2x + 1)(x - 1)(x² + x + 1)

 

 

Answer 2

 

24x⁸ - 21x⁵ - 3x² = 3x² (8x⁶ - 7x³ - 1) = 3x² (8x⁶ - 8x³ + 1x³ - 1) =

 

3x² [ 8x³(x³ - 1) + 1(x³ - 1)] = 3x² (x³ - 1) (2³x³ + 1³) =

 

3x² (x - 1)(x² + x + 1)(2x + 1)(4x² - 2x + 1)

Answer 3

24x⁸-21x⁵-3x²

the GCF is 3x²

3x²(8x⁶-7x³-1)

3x²(8x³+1)(x³-1)

8x³+1 is the sum of two cubes (a³+b³)=(a+b)(a²-ab+b²)

x³-1 is the difference of two cubes (a³-b³)=(a-b)(a²+ab+b²)

8x³+1=(2x+1)(4x²-2x+1)

x³-1=(x-1)(x²+x+1)

3x²(2x+1)(x-1)(4x²-2x+1)(x²+x+1) is the answer; you can't factor any more if you want integer coefficients