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.
The question that teacher has asked is that show that (3x-1)(x+5)(4x-3)=12x^3+47x^2-62x+15 for all values of x
To show that (3x-1)(x+5)(4x-3) is equal to 12x^3 + 47x^2 - 62x +15, we need to expand the expression and simplify it.
First, let's expand (3x-1)(x+5):
(3x-1)(x+5) = 3x(x+5) - 1(x+5) = 3x^2 + 15x - x - 5 = 3x^2 + 14x - 5
Now, let's substitute this into (3x-1)(x+5)(4x-3):
(3x-1)(x+5)(4x-3) = (3x^2 + 14x - 5)(4x-3)
To expand this, we apply the distributive property:
(3x^2 + 14x - 5)(4x-3) = 3x^2(4x-3) + 14x(4x-3) - 5(4x-3)
Now, we multiply each term:
= 12x^3 - 9x^2 + 56x^2 - 42x - 20x + 15
Combining like terms:
= 12x^3 + 47x^2 - 62x + 15
Therefore, (3x-1)(x+5)(4x-3) is equal to 12x^3 + 47x^2 - 62x + 15 for all values of x.
I'm confused
just expand the left side. It will be the same as the right side. One way to start would be
(3x-1)(x+5)(4x-3)
= 3x(x+5)(4x-3)-(x+5)(4x-3)
Now, (x+5)(4x-3)=4x^2+17x-15 so you can proceed with
= 3x(4x^2+17x-15)-(4x^2+17x-15)
= ...