Question

answer and get brainliest!!!!!

Answer 1

Answer: To prove they're zeroes:

Using the factor theorum, let us substitute x with -2 and 1/4 respectively.

1) 4* (-2)^3 - 5*(-2)^2 - 23*-2 + 6

= 4* (-8) - 5*(4) + 46 + 6

or -32 - 20 + 52 = 0

As the equation equals to 0 via substitution, -2 is a zero of the polynomial.

2) 4* (-1/4)^3 - 5*(-1/4)^2 - 23*-1/4 + 6

= 4* (1/64) - 5*(1/16) - 23/4 + 6

= 1/16 - 5/16 - 92/16 + 96/16

or 0

So, 1/4 is a zero as well.

To find the third zero, divide the polynomial by (x - 1/4) (x + 2) or $x^{2}$ + 7/4x - 1/2

Upon which you will get 4 (x - 3)

Meaning the third zero is 3 (Ans.)