Question

factor the following expressions
4(x+3)^2+ 8(x+3)-5

Answer 1

To factor the expression 4(x+3)^2 + 8(x+3) - 5, we can treat it as a quadratic trinomial in the form of ax^2 + bx + c and attempt to factor it.

Let's simplify the expression first:
4(x+3)^2 + 8(x+3) - 5
= 4(x^2 + 6x + 9) + 8(x+3) - 5
= 4x^2 + 24x + 36 + 8x + 24 - 5
= 4x^2 + 32x + 55

Now, let's try factoring the quadratic trinomial:
Since the coefficient of the x^2 term is 4, we need to find two binomials that multiply to give 4x^2. The only possibility is 2x and 2x.
Since the constant term is 55, we need to find two numbers that multiply to give 55 and add up to 32 (the coefficient of the x term). The numbers are 5 and 11.
Therefore, the factored form is:
4x^2 + 32x + 55 = (2x + 5)(2x + 11)

Answer 2

or, consider the quadratic 4u^2+8u+5 = (2u-1)(2u+5)
so, now we have
4(x+3)^2+ 8(x+3)-5 = (2(x+3)-1)(2(x+3)+5)
= (2x+5)(2x+11)

Answer 3

Apologies for the mistake in my previous response. You are correct. The correct factorization is:

4(x+3)^2 + 8(x+3) - 5 = (2(x+3)-1)(2(x+3)+5)
= (2x+5)(2x+11)