How do you factor the trinomial $2 a^{2} + 19 a - 10$ $2a^2+19a-10$ ?

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Answer 1 · 2016-07-14T10:36:23

$= (a + 10) (2 a - 1)$ $=(a+10)(2a-1)$

you can factor a trynomial

$a x^{2} + b x + c$ $ax^2+bx+c$

by applying the rule:

$a x^{2} + b x + c = a (x - x_{1}) (x - x_{2})$ $ax^2+bx+c=a(x-x_1)(x-x_2)$

where $x_{1} and x_{2}$ $x_1 and x_2$ are the zeroes of the trynomial

So, you first solve the equation:

$2 a^{2} + 19 a - 10 = 0$ $2a^2+19a-10=0$

$a = \frac{- 19 \pm \sqrt{19^{2} - 4 \cdot 2 \cdot (- 10)}}{2 \cdot 2}$ $a=(-19+-sqrt(19^2-4*2*(-10)))/(2*2)$

$a = \frac{- 19 \pm \sqrt{361 + 80}}{4}$ $a=(-19+-sqrt(361+80))/4$

$a = \frac{- 19 \pm \sqrt{441}}{4}$ $a=(-19+-sqrt(441))/4$

$a = \frac{- 19 \pm 21}{4}$ $a=(-19+-21)/4$

$a_{1} = - 10 and a_{2} = \frac{1}{2}$ $a_1=-10 and a_2=1/2$

Then

$2 a^{2} + 19 a - 10 = 0 = 2 (a + 10) (a - \frac{1}{2})$ $2a^2+19a-10=0=2(a+10)(a-1/2)$

$= (a + 10) (2 a - 1)$ $=(a+10)(2a-1)$

How do you factor the trinomial 2a2+19a−102a^2+19a-10?