How do you factor #2x^{2}+4x-6#?

2 Answers
anskat
Jun 11, 2018

You cannot (unless you use imaginary numbers, i.e. i).
Maybe check that there is no typo.

Explanation:

The Quadratic Formula: For #ax^2" + bx + c = 0#, the values of x (factors) which are the solutions of the equation are given by:
#x=(b+-sqrt(4*a*c))/(2*a)#

Here, a=2, b=4 and c=-6, so

#x=(4+-sqrt(4*2*(-6)))/(2*2)#

#x=(4+-sqrt(-48)/4)#

since you cannot take the square root of a negative number (-48), there are no real factors of teh equation. this means that, when plotted on a graph, the X-axis is never crossed.

Jim G.
Jun 11, 2018

#2(x+3)(x-1)#

Explanation:

#"take out a "color(blue)"common factor "2#

#=2(x^2+2x-3)#

#"the factors of - 3 which sum to + 2 are + 3 and - 1"#

#=2(x+3)(x-1)#

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