Factorize $2 (a + b) + \frac{a + b}{6} + {(a + b)}^{2} - 2$ $2(a+b)+(a+b)/6+(a+b)^2-2$ ?

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Factorize $2 (a + b) + \frac{a + b}{6} + {(a + b)}^{2} - 2$ $2(a+b)+(a+b)/6+(a+b)^2-2$ ?

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Answer 1 · 2017-03-13T07:07:40

$2 (a + b) + \frac{a + b}{6} + {(a + b)}^{2} - 2 = (a + b + \frac{13}{12} + \frac{\sqrt{457}}{12}) (a + b + \frac{13}{12} - \frac{\sqrt{457}}{12})$ $2(a+b)+(a+b)/6+(a+b)^2-2=(a+b+13/12+sqrt457/12)(a+b+13/12-sqrt457/12)$

Explanation:

$2 (a + b) + \frac{a + b}{6} + {(a + b)}^{2} - 2$ $2(a+b)+(a+b)/6+(a+b)^2-2$

= $(2 + \frac{1}{6}) (a + b) + {(a + b)}^{2} - 2$ $(2+1/6)(a+b)+(a+b)^2-2$

= $\frac{13}{6} (a + b) + {(a + b)}^{2} - 2$ $13/6(a+b)+(a+b)^2-2$

= ${(a + b)}^{2} + \frac{13}{6} (a + b) - 2$ $(a+b)^2+13/6(a+b)-2$

Let $a + b = x$ $a+b=x$ , then quadratic polynomial inside square bracket becomes $x^{2} + \frac{13}{6} x - 2$ $x^2+13/6x-2$ and as

$x^{2} + \frac{13}{6} x - 2$ $x^2+13/6x-2$

= $x^{2} + 2 \times \frac{13}{12} \times x + {(\frac{13}{12})}^{2} - 2 - {(\frac{13}{12})}^{2}$ $x^2+2xx13/12xx x+(13/12)^2-2-(13/12)^2$

= ${(x + \frac{13}{12})}^{2} - 2 - \frac{169}{144}$ $(x+13/12)^2-2-169/144$

= ${(x + \frac{13}{12})}^{2} - \frac{457}{144}$ $(x+13/12)^2-457/144$

= ${(x + \frac{13}{12})}^{2} - {(\frac{\sqrt{457}}{12})}^{2}$ $(x+13/12)^2-(sqrt457/12)^2$

= $(x + \frac{13}{12} + \frac{\sqrt{457}}{12}) (x + \frac{13}{12} - \frac{\sqrt{457}}{12})$ $(x+13/12+sqrt457/12)(x+13/12-sqrt457/12)$

and substituting $x$ $x$ with $a + b$ $a+b$ , we get

$2 (a + b) + \frac{a + b}{6} + {(a + b)}^{2} - 2 = (a + b + \frac{13}{12} + \frac{\sqrt{457}}{12}) (a + b + \frac{13}{12} - \frac{\sqrt{457}}{12})$ $2(a+b)+(a+b)/6+(a+b)^2-2=(a+b+13/12+sqrt457/12)(a+b+13/12-sqrt457/12)$

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Factorize $2 (a + b) + \frac{a + b}{6} + {(a + b)}^{2} - 2$ $2(a+b)+(a+b)/6+(a+b)^2-2$ ?

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Factorize $2 (a + b) + \frac{a + b}{6} + {(a + b)}^{2} - 2$ $2(a+b)+(a+b)/6+(a+b)^2-2$ ?