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

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

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Answer 1 · 2015-05-12T23:29:00

$2 x^{2} - 3 x - 2 = (2 x + 1) (x - 2)$ $2x^2 - 3x - 2 = (2x + 1)(x - 2)$

If a polynomial with integer coefficients has a rational root of the form $\frac{p}{q}$ $p/q$ in lowest terms then p must be a divisor of the constant term and q a divisor of the coefficient of the highest order term.

In your case any possible rational roots of $2 x^{2} - 3 x - 2 = 0$ $2x^2 - 3x - 2 = 0$ must be $\pm 2$ $+-2$ , $\pm 1$ $+-1$ or $\pm \frac{1}{2}$ $+-1/2$ . By trying these values you will find that $x = 2$ $x = 2$ and $x = - \frac{1}{2}$ $x = -1/2$ are zeros, so $(2 x + 1)$ $(2x + 1)$ and $(x - 2)$ $(x - 2)$ are factors.

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

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