How do you factor $3 x^{2} + 10 x - 8$ $3x^2 +10x - 8$ ?

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How do you factor $3 x^{2} + 10 x - 8$ $3x^2 +10x - 8$ ?

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Answer 1 · 2018-04-01T15:09:12

$(3 x - 2) (x + 4)$ $(3x-2)(x+4)$

Explanation:

$3 x^{2} + 10 x - 8$ $3x^2+10x-8$

first multiply $3$ $3$ with $- 8$ $-8$ and we get $- 24$ $-24$

then find two numbers that if we add or subtract we get $+ 10$ $+10$ and if we multiply we get $- 24$ $-24$

and the numbers are $+ 12$ $+12$ and $- 2$ $-2$

then we place those numbers as follow :-

$3 x^{2} + 12 x - 2 x - 8$ $3x^2+12x-2x-8$

then we group them

$(3 x^{2} + 12 x) (- 2 x - 8)$ $(3x^2+12x)(-2x-8)$

now take the common factor out.

$3 x (x + 4) - 2 (x + 4)$ $3x(x+4)-2(x+4)$

then group the numbers outside of the brackets

$(3 x - 2) (x + 4)$ $(3x-2)(x+4)$

and that is the answer

Proper Solution:-

$3 x^{2} + 10 x - 8$ $3x^2+10x-8$

$3 x^{2} + 12 x - 2 x - 8$ $3x^2+12x-2x-8$

$(3 x^{2} + 12 x) (- 2 x - 8)$ $(3x^2+12x)(-2x-8)$

$3 x (x + 4) - 2 (x + 4)$ $3x(x+4)-2(x+4)$

$(3 x - 2) (x + 4)$ $(3x-2)(x+4)$

Answer 2 · 2018-04-01T15:09:32

$(x + 4) (3 x - 2)$ $(x+4)(3x-2)$

Explanation:

$for a quadratic in standard form$ $"for a quadratic in "color(blue)"standard form"$

$∙ x a x^{2} + b x + c; a \neq 0$ $•color(white)(x)ax^2+bx+c ;a!=0$

$consider the factors of the product ac which sum to b$ $"consider the factors of the product ac which sum to b"$

$3 x^{2} + 10 x - 8 \leftarrow is in standard form$ $3x^2+10x-8larrcolor(blue)"is in standard form"$

$with a = 3, b = 10 and c = - 8$ $"with "a=3,b=10" and "c=-8$

$consider factors of 3 \times - 8 = - 24 which sum to + 10$ $"consider factors of "3xx-8=-24" which sum to + 10"$

$the required factors are + 12 and - 2$ $"the required factors are "+12" and "-2$

$split the middle term using these factors$ $color(blue)"split the middle term using these factors"$

$3 x^{2} + 12 x - 2 x - 8 \leftarrow factor by grouping$ $3x^2+12x-2x-8larrcolor(blue)"factor by grouping"$

$= 3 x (x + 4) - 2 (x + 4)$ $=color(red)(3x)(x+4)color(red)(-2)(x+4)$

$take out the common factor (x + 4)$ $"take out the common factor "(x+4)$

$= (x + 4) (3 x - 2)$ $=(x+4)(color(red)(3x-2))$

$\Rightarrow 3 x^{2} + 10 x - 8 = (x + 4) (3 x - 2)$ $rArr3x^2+10x-8=(x+4)(3x-2)$

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How do you factor $3 x^{2} + 10 x - 8$ $3x^2 +10x - 8$ ?

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How do you factor $3 x^{2} + 10 x - 8$ $3x^2 +10x - 8$ ?