How do you factor 24x^2-76x+40?

3 Answers
May 15, 2015

First, let's simplify this dividing the equation by 4.

6x^2-19x+10

Then, we have to find its roots: x_1=2/3 and x_2=5/2. But in order not to work with these fractions, we can multiply both sides by the denominator of the fractions, leaving only rational numbers:

3x_1=2/cancel(3)*cancel(3) which means 3x_1-2=0
2x_1=5/cancel(2)*cancel(2) which means 2x_2-5=0

Now, let's just factor these.

(3x-2)(2x-5)

Done.

May 15, 2015

24x^2-76x+40 = 4*6x^2-4*19x+4*10

= 4*(6x^2-19x+10)

From the rational root theorem, if p/q is a root of 6x^2-19x+10 = 0 written in lowest terms, then p must be a divisor of the constant term 10 and q must be a divisor of the coefficient (6) of the highest order term.

Notice that 6 can be factored as 2*3 and 10 can be factored as 2*5.

Let us try (2x-5)(3x-2), it seems to work...

(2x-5)(3x-2) = 2x*3x-5*3x-2x*2-5*(-2)

= 6x^2 - (15+4)x + 10

= 6x^2 - 19x + 10

So in summary, 24x^2-76x+40 = 4*(2x-5)(3x-1)

May 15, 2015

24x^2-76x+40 = 4(6x^2-19x+10).

To factor 6x^2-19x+10 multiply 6xx10 = 60.

We need two numbers that multiply to give us 60 and add to get us -19 (the number in the middle).

A little thought should convince us that both numbers must be negative. We proceed in order:

-1xx-60 do not add up to -19

-2xx-30 do not add up to -19

-3xx-10 do not add up to -19

-4xx-15 STOP! These do add up to -19.

We'll split the -19x into -4x-15x (or -15x-4x either will work)

6x^2-19x+10 = 6x^2-4x-15x+10.

Now factor by grouping:

6x^2-19x+10 = (6x^2-4x)+(-15x+10).

6x^2-19x+10 = 2x(3x-3)+(-5)(3x-2) .

6x^2-19x+10 = (2x-5)(3x-2).

So we finish with:

24x^2-76x+40 = 4(6x^2-19x+10)

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