How do you factor #3x^2-2x-8#?

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Answer 1 · 2015-02-03T17:35:02

The solution is: #(3x+4)(x-2)#

The polynomial is of the type #ax^2+bx+c#.

There are two ways to factor.

The first one:

we have to find two numbers whose sum is #-2# (that is #c#) and whose product is #-24# (that is #a*c#).

#-24=-1*24#
#-24=-24*1#
#-24=-2*12#
#-24=-12*2#
#-24=-3*8#
#-24=-8*3#
#-24=-4*6#
#-24=-6*4#

The couple of numbers whose sum is #-2# is #-6,4#.

Now we have to "split" the monomial #-2x# in two parts: #-2x=-6x+4x#.

So our polynomial becomes:

#3x^2-6x+4x-8=3x(x-2)+4(x-2)=(x-2)(3x+4)#.

If it is known the way to solve an equation of 2° degree, here there is the second way to factor it:

We can imagine that this polynomial "becomes" an equation:

#3x^2-2x-8=0#, we can calculate the quantity

#Delta=b^2-4ac=4+96=100#,

and than we have to use this formula:

#x_(1,2)=(-b+-sqrtDelta)/(2a)=(2+-10)/6rArrx_1=-4/3;x_2=2#.

And now we have to use the formula:

#ax^2+bx+c=a(x-x_1)(x-x_2)#.

So:

#3x^2-2x-8=3(x+4/3)(x-2)=(3x+4)(x-2)#