How do you factor $9 x {(3 x - 9)}^{2} + {(3 x - 9)}^{3}$ $9x(3x-9)^2 + (3x-9)^3$ ?

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How do you factor $9 x {(3 x - 9)}^{2} + {(3 x - 9)}^{3}$ $9x(3x-9)^2 + (3x-9)^3$ ?

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Answer 1 · 2015-11-22T16:25:57

$= 3 (4 x - 3) {(3 x - 9)}^{2}$ $=3(4x-3)(3x-9)^2$

Explanation:

$9 x {(3 x - 9)}^{2} + {(3 x - 9)}^{3} = (9 x + 3 x - 9) {(3 x - 9)}^{2}$ $9x(3x-9)^2+(3x-9)^3 = (9x + 3x -9)(3x-9)^2$

$= (12 x - 9) {(3 x - 9)}^{2}$ $=(12x - 9)(3x-9)^2$

$= 3 (4 x - 3) {(3 x - 9)}^{2}$ $=3(4x-3)(3x-9)^2$

You know that

${(3 x - 9)}^{2} = {[3 (x - 3)]}^{2} = 9 \cdot {(x - 3)}^{2}$ $(3x-9)^2= [3(x-3)]^2 = 9 * (x-3)^2$

The expression becomes

$3 (4 x - 3) {(3 x - 9)}^{2} = 27 \cdot (4 x - 3) {(x - 3)}^{2}$ $3(4x-3)(3x-9)^2 = 27 * (4x-3)(x-3)^2$

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How do you factor $9 x {(3 x - 9)}^{2} + {(3 x - 9)}^{3}$ $9x(3x-9)^2 + (3x-9)^3$ ?

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Explanation:

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How do you factor $9 x {(3 x - 9)}^{2} + {(3 x - 9)}^{3}$ $9x(3x-9)^2 + (3x-9)^3$ ?