Factor completely: (x-9)(3x+5)+(4x+9)(3x+5)^(2) Answer:

Question

Factor completely:  (x-9)(3x+5)+(4x+9)(3x+5)^(2) Answer:

Bytelearn · Accepted Answer

Identify common factors: Identify common factors in both terms of the expression. The expression is (x-9)(3x+5)+(4x+9)(3x+5)^(2). We can see that (3x+5) is a common factor.Factor out common factor: Factor out the common factor (3x+5). We can write the expression as (3x+5)[(x-9) + (4x+9)(3x+5)].Distribute inside the brackets: Distribute the (3x+5) inside the brackets. Now we need to multiply (4x+9) by (3x+5) and add (x-9) to it. This gives us (3x+5)[(x-9) + (12x^2 + 27x + 45)].Combine like terms: Combine like terms inside the brackets. We combine (x-9) with (12x^2 + 27x + 45) to get (3x+5)(12x^2 + 28x + 36).Check for further factoring: Check for any further factoring possibilities. The quadratic expression 12x^2 + 28x + 36 can be factored further. We look for two numbers that multiply to 12*36 and add to 28. These numbers are 12 and 24.Factor quadratic expression: Factor the quadratic expression. We can write 12x^2 + 28x + 36 as (4x+6)(3x+6). So the expression becomes (3x+5)(4x+6)(3x+6).Check for common factors: Check for any common factors or simplifications. There are no common factors between (3x+5), (4x+6), and (3x+6), and no further simplifications can be made.

Factor completely: $\newline$ $(x-9)(3 x+5)+(4 x+9)(3 x+5)^{2}$ $\newline$ Answer:

Full solution

More problems from Powers with negative bases

Related topics

Factor completely:\newline(x−9)(3x+5)+(4x+9)(3x+5)2 (x-9)(3 x+5)+(4 x+9)(3 x+5)^{2} (x−9)(3x+5)+(4x+9)(3x+5)2\newlineAnswer:

Factor completely: $\newline$ $(x-9)(3 x+5)+(4 x+9)(3 x+5)^{2}$ $\newline$ Answer: