How do you factor $x^{2} - 7 x b + 6 b^{2}$ $x^2 - 7xb + 6b^2$ ?

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Answer 1 · 2015-06-08T06:48:13

Since all the terms are of order $2$ $2$ it's similar to factoring a quadratic in one variable $x^{2} - 7 x + 6$ $x^2-7x+6$ .

$x^{2} - 7 x + 6 = (x - 1) (x - 6)$ $x^2-7x+6=(x-1)(x-6)$

$1 + 6 = 7$ $1 + 6 = 7$ and $1 \times 6 = 6$ $1 xx 6 = 6$

$x^{2} - 7 x b + 6 b^{2} = (x - b) (x - 6 b)$ $x^2-7xb+6b^2 = (x-b)(x-6b)$

How do you factor x^2 - 7xb + 6b^2x2−7xb+6b2x^2 - 7xb + 6b^2?