how do you factorise 2x^2 -9x+4?

Answer 1

2x^2 - 9x + 4 = 2(x^2 - 4.5x + 2) =
= 2(x^2 - 2*2.25x + 2.25^2 - 2.25^2 + 2) =
= 2[(x - 2.25)^2 - 5.0625 + 2] =
= 2[(x - 2.25)^2 - 3.0625] =
= 2[(x - 2.25)^2 - 1.75^2] =
= 2(x - 2.25 - 1.75)(x - 2.25 + 1.75) =
= 2(x - 4)(x - 0.5)

Answer 2

Hi,

The best approach (most times) is to list out all of the UNIQUE the possible combinations (while leaving out the sign). In this case:

(2x 2)(x 2)
(2x 4)(x 1)
(2x 1)(x 4)

Now realize what you want to end up with for the middle term. Since in the first and second options you'd have 6x, and we need to have -9x, those are out.

Now you know that you need the third one, and all you have to do is select the correct signs. Since you need to end up with -9x, and the last term is +4, they both need to be negative.

So you end up with
(2x -1)(x-4)

QED

Hope that helps,
Matt

Answer 3

Rough Work:
1. Multiply 2x^2 and 4 (2x^2 * 4 = 8x^2)
2. Now Choose the factors whose sum comes out to be the middle term (-9x) and product is equal to 8x^2
3. Hmmmm.. now we can choose -8x and -x (sum = -9x & Product = +8x^2)

Neat Work:
Step 1. Write the equation as
2x^2 - 8x - x + 4
Step 2. Now take 2x common in first half n 1 in the second
2x(x-4) - 1(x-4)
Step 3. the Answer is
(2x-1)(x-4)

Check:
Multiply again and u'll get the answer

Answer 4

2x^2-9x+4
middle term breaking
2x^2-8x-x+4
selecting and separating common term
2x(x-4) -1(x-4)
selecting and separating common term again
(2x-1) (x-4). these are your roots. both are real

Answer 5

(2x-1)(x-4)

Answer 6

(2x-1)(x-4)