Question

Show that: (3x-1)(x+5)(4x-3) = 12x^3 + 47x^2 - 62x +15 for all values of x. (3x-1)(x+5)(4x-3) = (?????)(4x-3) Expand into 6 terms: ???????? Simplify: ???????? Thank you.

Answer 1

(3x - 1)( x + 5)(4x - 3) = 12x^3 + 47x^2 - 62x +15 
(3x^2 – x + 15x - 5)( 4x – 3 ) = 12x^3 + 47x^2 - 62x +15 
(3x^2 + 14x - 5)( 4x – 3 ) = 12x^3 + 47x^2 - 62x +15 
12x^3 – 9x^2 + 56x^2 – 42x – 20x + 15 = 12x^3 + 47x^2 - 62x +15 
12x^3 – 47x^2 - 62x + 15 = 12x^3 + 47x^2 - 62x +15 

 

Answer 2

Answer with Step-by-step explanation:

We have to show that:

$(3x-1)(x+5)(4x-3) = 12x^3 + 47x^2 - 62x +15$

$(3x-1)(x+5)(4x-3) = (3x(x+5)-1(x+5))(4x-3)\\\\=(3x^2+15x-x-5)(4x-3)\\\\=(3x^2+14x-5)(4x-3)\\\\=3x^2(4x-3)+14x(4x-3)-5(4x-3)\\\\=12x^3-9x^2+56x^2-42x-20x+15\\\\=12x^3+47x^2-62x+15$

Hence, we have

$(3x-1)(x+5)(4x-3) = 12x^3 + 47x^2 - 62x +15$