Answer:
z = 9 will allow both polynomials to be factored.
Step-by-step explanation:
Perfect square binomials:
For perfect square binomials, we have that:
[tex](x \pm a)^2 = x^2 \pm 2ax + a^2[/tex]
x² - 6x + z
x² + 6x + z
First term of the binomial is x, and the second is:
[tex]2ax = \pm 6x[/tex]
[tex]2a = \pm 6[/tex]
[tex]a = \pm \frac{6}{2}[/tex]
[tex]a = \pm 3[/tex]
Value of z:
The final term is a squared, so:
[tex]a = (\pm 3)^2 = 9[/tex]
z = 9 will allow both polynomials to be factored.
