How do you factor the trinomial $3 x^{2} - 15 x + 16$ $3x^2-15x+16$ ?

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How do you factor the trinomial $3 x^{2} - 15 x + 16$ $3x^2-15x+16$ ?

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Answer 1 · 2016-03-07T08:44:57

Factors are $3 (x - \frac{15 + \sqrt{33}}{6}) (x - \frac{15 - \sqrt{33}}{6})$ $3(x-(15+sqrt33)/6)(x-(15-sqrt33)/6)$

Explanation:

To factorize $a x^{2} + b x + c$ $ax^2+bx+c$ , one needs to first check about discriminant $b^{2} - 4 a c$ $b^2-4ac$ , which in $3 x^{2} - 15 x + 16$ $3x^2−15x+16$ is ${(- 15)}^{2} - 4 \times 3 \times 16 = 225 - 192 = 33$ $(-15)^2-4xx3xx16=225-192=33$ . As it is not the square of a rational number, we will not have binomials with rational coefficients as factors.

Hence, to find factors let us solve the equation $3 x^{2} - 15 x + 16 = 0$ $3x^2−15x+16=0$ using quadratic formula which gives solution of $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
as $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$ .

Hence solution of $3 x^{2} - 15 x + 16 = 0$ $3x^2−15x+16=0$ is $x = \frac{- (- 15) \pm \sqrt{{(- 15)}^{2} - 4 \times 3 \times 16}}{2 \times 3}$ $x=(-(-15)+-sqrt((-15)^2-4xx3xx16))/(2xx3)$ or

$x = \frac{15 \pm \sqrt{33}}{6}$ $x=(15+-sqrt33)/6$

Hence factors are $3 (x - \frac{15 + \sqrt{33}}{6}) (x - \frac{15 - \sqrt{33}}{6})$ $3(x-(15+sqrt33)/6)(x-(15-sqrt33)/6)$

We have multiplied by $3$ $3$ as coefficient of $x^{2}$ $x^2$ is $3$ $3$ .

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How do you factor the trinomial $3 x^{2} - 15 x + 16$ $3x^2-15x+16$ ?

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