5x^2-9x+4=0?

Use greek method of facotrization:

Multiply 5*4 = 20

Find two factors of 20 that sum to -9.

Those factors are -5 and -4

5x² + -5x - 4x + 4 = 0

(5x² - 5x) + (-4x + 4) = 0

5x(x - 1) - 4(x - 1) = 0

(x - 1)(5x - 4) = 0

x - 1 = 0 or 5x - 4 = 0

x = 1 or x = 4/5

(5x - 4)(x - 1) = 0
x = 4 / 5 , x = 1

x = 4/5 and x = 1

This can factor to:

( 5x - 4 )(x - 1)

To find the numbers, use reverse FOIL:
First of all, since the last sign is positive and the first is negative, this means the factorizations must be of form

(mx - k)(nx - J)

Multiply the first coefficient with the last (5 * 4) = 20

Find the factors of 20: 20,1: 10,2: 4,5

Which two add up to the middle coefficient (9)? It must be 5 and 4. With a little juggling, you can see exactly where they fit.

Then just set each factor to zero to find the solutions:

5x - 4 = 0
x = 4/5

x - 1 = 0
x = 1

Use the quadratic formula here, since that method turns out to be quite simple.

x+(-b+-sqrt(b^2-4ac))/2a

Note that 9^2-4*4*5=1, so that simplifies things considerably.

Now we just have (9+/-1)/10, so x is 1 or 4/5.

This is a quadratic that can be solved either by factoring or the quadratic formula

It factors as

(5x - 4)(x - 1) = 0

so

5x - 4 = 0 -> 5x = 4 -> x = 4/5

x - 1 = 0 -> x = 1

5x^2 - 9x + 4 = 0
(5x - 4)(x - 1) = 0

5x - 4 = 0
5x = 4
x = 4/5 (0.8)

x - 1 = 0
x = 1

∴ x = 4/5 (0.8) , 1

5x^2 - 9x + 4 = 0
(5x - 4)(x -1) = 0
5x - 4 = 0 or x - 1 = 0
x = 4/5 or x = 1

Factor first.
(5x -4)(x-1) = 0
5x-4=0
5x=4
x= 5/4

x-1=0
x=1
Answers are x = 5/4 and 1

Factor the polynomial
(5x - 4)(x - 1) = 0

Set both factors equal to 0
5x - 4 = 0
x - 1 = 0

Solve for x
5x = 4
x = 5/4

x - 1 = 0
x = 1

The two solutions are x = 5/4 and x = 1