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

Answer:

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

Step-by-step 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

$-20x^{2}+23x-6=-20(x-(3/4))(x-(2/5))$

$-20x^{2}+23x-6=-(5)(4)(x-(3/4))(x-(2/5))$

$-20x^{2}+23x-6=-(4x-3)(5x-2)$