How do you factor the trinomial $x^{2} + 3 x - 36 = 0$ $x^2+3x-36=0$ ?

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How do you factor the trinomial $x^{2} + 3 x - 36 = 0$ $x^2+3x-36=0$ ?

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Answer 1 · 2015-12-26T13:03:35

$x^{2} + 3 x - 36 = 0$ $x^2+3x-36=0$ can be solved by quadratic formula but cannot be solved by integer factors. In case, you are looking for a solution it is explained below.

Explanation:

Quadratic equations of for $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ can be solved by the quadratic formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

Our question $x^{2} + 3 x - 36 = 0$ $x^2+3x-36 =0$
Compare with $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

$a = 1$ $a=1$ , $b = 3$ $b=3$ and $c = - 36$ $c=-36$

Substitute these values of $a$ $a$ , $b$ $b$ and $c$ $c$ in the formula

$x = \frac{- (3) \pm \sqrt{{(3)}^{2} - 4 (1) (- 36)}}{2 (1)}$ $x =(-(3)+-sqrt((3)^2-4(1)(-36)))/(2(1))$

$x = \frac{- 3 \pm \sqrt{9 + 144}}{2}$ $x=(-3+-sqrt(9+ 144))/2$

$x = \frac{- 3 \pm \sqrt{153}}{2}$ $x=(-3+-sqrt(153))/2$

$x = \frac{- 3 \pm \sqrt{3^{2} \cdot 17}}{2}$ $x=(-3+-sqrt(3^2*17))/2$
$x = \frac{- 3 \pm 3 \sqrt{17}}{2}$ $x=(-3+-3sqrt(17))/2$

Solutions:

$x = \frac{- 3 - 3 \sqrt{17}}{2}$ $x=(-3-3sqrt(17))/2$ or $x = \frac{- 3 + 3 \sqrt{17}}{2}$ $x=(-3+3sqrt(17))/2$

Real (non-integer) Factors
$x^{2} + 3 x - 36 = (x + \frac{3 - 3 \sqrt{17}}{2}) (x + \frac{3 + 3 \sqrt{17}}{2})$ $x^2+3x-36 = (x+(3-3sqrt(17))/2)(x+(3+3sqrt(17))/2)$

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How do you factor the trinomial $x^{2} + 3 x - 36 = 0$ $x^2+3x-36=0$ ?

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Explanation:

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How do you factor the trinomial x2+3x−36=0x^2+3x-36=0?

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Explanation:

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How do you factor the trinomial $x^{2} + 3 x - 36 = 0$ $x^2+3x-36=0$ ?