How do you factor 2x^2+5x-6?

1 Answer
Oct 10, 2015

2x^2+5x-6=2(x+5/4-sqrt(73)/4)(x+5/4+sqrt(73)/4)

Explanation:

Quadratic formula states that for an equation of the form ax^2 + bx + c=0,

x=(-b+sqrt(b^2-4ac))/(2a), (-b-sqrt(b^2-4ac))/(2a).

These two numbers are called roots.

Since plugging these 2 roots to ax^2 + bx + c gives 0, we can say that (x-root1)(x-root2)=0.

Thus, finding these roots means factoring the equation!
In this problem, a=2, b=5, and c=-6.

So

x_1=(-5+sqrt(5^2-4(2)(-6)))/(2(2)), x_2=(-5-sqrt(5^2-4(2)(-6)))/(2(2)).

x_1=(-5+sqrt(73))/(4), x_2=(-5-sqrt(73))/(4).

In conclusion,

2x^2+5x-6=2(x+5/4-sqrt(73)/4)(x+5/4+sqrt(73)/4).

I multiplied the whole factored equation by 2 because the factored equation starts with x^2 and the original equation has 2x^2.

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