Verified answer

It is not possible to factorize the expression by grouping in pairs.

Have you stated the question correctly ? Please check up. Is the middle term - 5w or - 8w ?

This can be factorized as follows :

12w² - 5w - 4

= 12{w² - (5/12)w - 1/3}

= 12{w² - 2(w)(5/24) + (5/24)² - (5/24)² - 1/3}

= 12{(w - 5/24)² - (25/576 + 1/3)}

= 12{(w - 5/24)² - (217/576)}

= 12{(w - 5/24)² - (√217/24)²}

= 12{(w - 5/24 + √217/24)(w - 5/24 - √217/24)}

= 12{w + (√217 - 5)/24}{w - (√217 + 5)}

This can't be factorised. In order to factorise, the middle-term breaking method is used, and according to that, the middle-term should be broken into two terms, whose sum is the middle term, while their product is the product of the first and last terms.

In this case, the product of the first and last terms-i.e 12 x 4-is 48

Now, write down the factors of 48

1 x 48

2 x 24

3 x 16

4 x 12

6 x 8

The sum/difference of none of these pair of terms is equal to 5; thus, this expression can neither be factorised, nor be simplified.

However, if it was an equation, you could solve it by either the completing square method, or the quadratic formula!

I'm editing my comment, to let you know that there must be a typo/mis-print with the question then!

Cause (3w-4)(2w+1) are the roots of 6w^2-5w-4

Considering it was 6w^2 instead of 12w^2, here's how you do it:

6w^2 - 5w - 4

The product of the first and last termz is 6 x (-4)=-24

The only two numbers that will add up to -5 and multiply to give the product of -24 are (-8) and (+3)

So, you break-up the middle term, and write it into two parts, like this:

6w^2 - 8w + 3w - 4

Now, factorise:

2w (3w-4) + 1 (3w-4)

Simplify:

(2w+1)(3w-4)

There ya go!

I hope that cleared out your confusion!

:)

Contrary to a previous answerer, the zeros of the function represented by this are not imaginary, but this doesn't factor; it won't break down into polynomial factoring, which requires integers. If you put A, B and C into the Quadratic Formula, you'll see that the discriminant is positive but ends in 7, and no perfect square does that.

question quantity a million : For this equation 2*x^2 + 7*x - 4 = 0 , answer right here questions : A. Use factorization to discover the basis of the equation ! answer quantity a million : The equation 2*x^2 + 7*x - 4 = 0 is already in a*x^2+b*x+c=0 variety. if you desire to mean that the fee of a = 2, b = 7, c = -4. 1A. Use factorization to discover the basis of the equation ! 2*x^2 + 7*x - 4 = 0 <=> ( 2*x - a million ) * ( x + 4 ) = 0

12w² - 5w - 4

use quadratic equations

(5+-sqrt(25-4*12*-4))/24

=(5+-sqrt(25+192))/24

=(5+-sqrt(217)))/24

roots are imaginary for the above eq