Factor completely: (2x-5)^(2)(x-5)-(2x-5)(5x+7) Answer:

Factor out common term: First, notice that the term (2x-5) is common in both parts of the expression. We can factor this term out.Factor out (2x-5): After factoring out (2x-5), the expression becomes: (2x-5)[(2x-5)(x-5)-(5x+7)]Distribute inside brackets: Now, distribute (2x-5) inside the brackets to the terms (x-5) and -(5x+7): (2x-5)[(2x-5)(x-5) - (5x+7)] = (2x-5)[2x(x-5) - 5(x-5) - (5x+7)]Distribute 2x and -5: Next, distribute 2x to (x-5) and -5 to (x-5): (2x-5)[2x^2 - 10x - 5x + 25 - 5x - 7]Combine like terms: Combine like terms within the brackets: (2x-5)[2x^2 - 20x + 18]Final fully factored expression: The expression is now fully factored: (2x-5)(2x^2 - 20x + 18)

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Factor completely:

(2x-5)^(2)(x-5)-(2x-5)(5x+7)
Answer:

Factor completely: $\newline$ $(2 x-5)^{2}(x-5)-(2 x-5)(5 x+7)$ $\newline$ Answer:

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Math Problems

Algebra 2

Sum of finite series starts from 1

Full solution

Q. Factor completely:

\newline

(2 x-5)^{2}(x-5)-(2 x-5)(5 x+7)

\newline

Answer:

Factor out common term: First, notice that the term $2x-5$ is common in both parts of the expression. We can factor this term out.
Factor out $(2x-5)$ : After factoring out $(2x-5)$ , the expression becomes: $(2x-5)[(2x-5)(x-5)-(5x+7)]$
Distribute inside brackets: Now, distribute $(2x-5)$ inside the brackets to the terms $(x-5)$ and $-(5x+7)$ : $(2x-5)[(2x-5)(x-5) - (5x+7)]$ = $(2x-5)[2x(x-5) - 5(x-5) - (5x+7)]$
Distribute $2x$ and $-5$ : Next, distribute $2x$ to $(x-5)$ and $-5$ to $(x-5)$ : $(2x-5)[2x^2 - 10x - 5x + 25 - 5x - 7]$
Combine like terms: Combine like terms within the brackets: \(2x $-5$ )\left[ $2$ x^ $2$ - $20$ x + $18$ \right]
Final fully factored expression: The expression is now fully factored: \(2x $-5$ )( $2$ x^ $2$ - $20$ x + $18$ )\