Factor completely: (x+1)^(2)-(5x-8)^(2) Answer:

Recognize as difference of squares: Recognize the expression as a difference of squares. The given expression is in the form of a^2 - b^2, which is a difference of squares. The difference of squares can be factored into (a + b)(a - b).Identify a and b: Identify a and b for the difference of squares. In the expression (x+1)^(2)-(5x-8)^(2), a is (x+1) and b is (5x-8).Apply formula: Apply the difference of squares formula. Using the formula (a + b)(a - b), we can write the factored form of the expression as: ((x+1) + (5x-8))((x+1) - (5x-8))Simplify each factor: Simplify each factor. First factor: (x+1) + (5x-8) = x + 1 + 5x - 8 = 6x - 7 Second factor: (x+1) - (5x-8) = x + 1 - 5x + 8 = -4x + 9Write final form: Write the final factored form. The factored form of the expression is (6x - 7)(-4x + 9).

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Factor completely:

(x+1)^(2)-(5x-8)^(2)
Answer:

Factor completely: $\newline$ $(x+1)^{2}-(5 x-8)^{2}$ $\newline$ Answer:

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Math Problems

Grade 6

Factor numerical expressions using the distributive property

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Q. Factor completely:

\newline

(x+1)^{2}-(5 x-8)^{2}

\newline

Answer:

Recognize as difference of squares: Recognize the expression as a difference of squares. The given expression is in the form of $a^2 - b^2$ , which is a difference of squares. The difference of squares can be factored into $(a + b)(a - b)$ .
Identify $a$ and $b$ : Identify $a$ and $b$ for the difference of squares. $\newline$ In the expression $(x+1)^{2}-(5x-8)^{2}$ , $a$ is $(x+1)$ and $b$ is $(5x-8)$ .
Apply formula: Apply the difference of squares formula. $\newline$ Using the formula $(a + b)(a - b)$ , we can write the factored form of the expression as: $\newline$ $((x+1) + (5x-8))((x+1) - (5x-8))$
Simplify each factor: Simplify each factor. $\newline$ First factor: $(x+1) + (5x-8) = x + 1 + 5x - 8 = 6x - 7$ $\newline$ Second factor: $(x+1) - (5x-8) = x + 1 - 5x + 8 = -4x + 9$
Write final form: Write the final factored form. $\newline$ The factored form of the expression is $(6x - 7)(-4x + 9)$ .