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

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Answer 1 · 2015-10-10T22:18:14

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

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#.