How do you factor $7 x^{2} - 9 x + 2$ $7x^2 - 9x + 2$ ?

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How do you factor $7 x^{2} - 9 x + 2$ $7x^2 - 9x + 2$ ?

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Answer 1 · 2015-06-06T16:54:48

You find its roots and turn them into factors by equaling them to zero.

Let's find the roots using Bhaskara:

$\frac{9 \pm \sqrt{81 - 4 (7) (2)}}{14}$ $(9+-sqrt(81-4(7)(2)))/14$

$\frac{9 \pm \sqrt{81 - 56}}{14}$ $(9+-sqrt(81-56))/14$

$\frac{9 \pm \sqrt{25}}{14}$ $(9+-sqrt25)/14$

$\frac{9 \pm 5}{14}$ $(9+-5)/14$

$x_{1} = 1$ $x_1=1$ , which, equaled to zero is $x - 1 = 0$ $x-1=0$
$x_{2} = \frac{2}{7}$ $x_2=2/7$ , which, equaled to zero is $7 x - 2 = 0$ $7x-2=0$

Thus, we can factor $7 x^{2} - 9 x + 2$ $7x^2-9x+2$ this way:

$(x - 1) (7 x - 2)$ $(x-1)(7x-2)$

Answer 2 · 2015-06-06T21:56:32

$f (x) = 7 x^{2} - 9 x + 2 .$ $f(x) = 7x^2 - 9x + 2.$

Since (a + b + c) = 0, one factor is (x - 1) and the other is $(- \frac{c}{a} = - \frac{2}{7})$ $(-c/a = -2/7)$

f(x) = (x - 1)(7x - 2)

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How do you factor $7 x^{2} - 9 x + 2$ $7x^2 - 9x + 2$ ?

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