How do you factor $x^{3} + 7 x^{2} + 14 x + 8$ $x^3+7x^2+14x+8$ ?

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How do you factor $x^{3} + 7 x^{2} + 14 x + 8$ $x^3+7x^2+14x+8$ ?

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Answer 1 · 2018-03-05T18:16:28

$= (x + 1) (x + 4) (x + 2)$ $color(magenta)(=(x+1)(x+4)(x+2)$

Explanation:

$x^{3} + 7 x^{2} + 14 x + 8$ $x^3+7x^2+14x+8$

$= x^{3} - x + 7 x^{2} + 15 x + 8$ $=x^3-x+7x^2+15x+8$ [Adding and subtracting $x$ $x$ ]

$= x (x^{2} - 1) + 7 x^{2} + 15 x + 8$ $=x(x^2-1)+7x^2+15x+8$

Identity:
$a^{2} - b^{2} = (a + b) (a - b)$ $color(red)(a^2-b^2=(a+b)(a-b)$ &
$(x + a) (x + b) = x^{2} + (a + b) x + a b$ $color(red)((x+a)(x+b)=x^2+(a+b)x+ab$

$= x (x + 1) (x - 1) + [7 x (x + 1) + 8 (x + 1)$ $= x(x+1)(x-1)+[7x(x+1)+8(x+1)$

$= x (x + 1) (x - 1) + (x + 1) (7 x + 8)$ $= x(x+1)(x-1)+(x+1)(7x+8)$

$= (x + 1) [x (x - 1) (7 x + 8)]$ $= (x+1)[x(x-1)(7x+8)]$

$= (x + 1) [x^{2} - x + 7 x + 8]$ $= (x+1)[x^2-x+7x+8]$

$= (x + 1) [x^{2} + 6 x + 8]$ $= (x+1)[x^2+6x+8]$

Identity:
$(x + a) (x + b) = x^{2} + (a + b) x + a b$ $color(red)((x+a)(x+b)=x^2+(a+b)x+ab$

$= (x + 1) (x + 4) (x + 2)$ $color(magenta)(=(x+1)(x+4)(x+2)$

~Hope this helps! :)

Answer 2 · 2018-03-05T18:25:15

(x + 1)(x + 2)(x + 4)

Explanation:

$f (x) = x^{3} + 7 x^{2} + 14 x + 8$ $f(x) = x^3 + 7x^2 + 14x + 8$
First, we note that f(-1) = - 1 + 7 - 14 + 8 = 0.
So, one factor is (x - 1).
After division, or guest -->
$f (x) = (x + 1) (x^{2} + 6 x + 8)$ $f(x) = (x + 1)(x^2 + 6x + 8)$
Find 2 numbers knowing the sum (6) and the product (8). They are 2 and 4. Finally,
f(x) = (x + 1)(x + 2)(x + 4)

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