Question

NEED HELP!!!!!!!!!!!!!!!

Which factorizations can be used to identify the real zeros of the function f(x)=-20x^2+23x-6 ?

A. (-10x+2)(2x+3)
B. -(10x+2)(2x-3)
C. -(4x-3)(5x+2)
D. -(4x-3)(5x-2)

Answer 1

It's D, it's the only one where it has a -6 at the end other than A, but if you look at the A when distributed there isn't a 23x

Answer 2

Option D.
`-(4x-3)(5x-2)`

Explanation:

we have

`f(x)=-20x^(2)+23x-6`

Equate the function to zero

`-20x^(2)+23x-6=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`-20x^(2)+23x=6`

Factor the leading coefficient

`-20(x^(2)-(23/20)x)=6`

Complete the square. Remember to balance the equation by adding the same constants to each side

`-20(x^(2)-(23/20)x+(529/1,600))=6-(529/80)`

`-20(x^(2)-(23/20)x+(529/1,600))=-(49/80)`

`(x^(2)-(23/20)x+(529/1,600))=(49/1,600)`

Rewrite as perfect squares

`(x-(23/40))^(2)=(49/1,600)`

`(x-(23/40))=(+/-)(7/40)`

`x=(23/40)(+/-)(7/40)`

`x=(23/40)(+)(7/40)=30/40=3/4`

`x=(23/40)(-)(7/40)=16/40=2/5`

therefore

`-20x^(2)+23x-6=-20(x-(3/4))(x-(2/5))`

`-20x^(2)+23x-6=-(5)(4)(x-(3/4))(x-(2/5))`

`-20x^(2)+23x-6=-(4x-3)(5x-2)`