Favor out a-1 as long as you do it to the whole problem it is now equivalent then just worry about factoring the new product
-1(8x^2+X-9)
(8x+9)((X-1)
8x+9=0
Subtract 9 divide by 8
X=-9/ 8 answer
X-1=0 X=1 answer
I believe the first is right might be wrong
I believe that the answer is A.
Answer:
b
Step-by-step explanation:
cause it a circle and it will need a little more then the number it is
Answer:
![P(x)=x^{4}-6x^{3}-16x^{2}+96x](https://tex.z-dn.net/?f=P%28x%29%3Dx%5E%7B4%7D-6x%5E%7B3%7D-16x%5E%7B2%7D%2B96x)
Step-by-step explanation:
1. The polynomial must have the following zeros:
![x=-4\\x=0\\x=4\\x=6](https://tex.z-dn.net/?f=x%3D-4%5C%5Cx%3D0%5C%5Cx%3D4%5C%5Cx%3D6)
2. This means the following:
![x+4=0\\x=0\\x-4=0\\x-6=0](https://tex.z-dn.net/?f=x%2B4%3D0%5C%5Cx%3D0%5C%5Cx-4%3D0%5C%5Cx-6%3D0)
3. Multiply each term. The product of the multiplicaction is equal to zero. Then:
![(x+4)(x)(x-4)(x-6)=0](https://tex.z-dn.net/?f=%28x%2B4%29%28x%29%28x-4%29%28x-6%29%3D0)
4.
and
are conjugates, therefore, you have:
![(x^{2}-16)(x)(x-6)=0](https://tex.z-dn.net/?f=%28x%5E%7B2%7D-16%29%28x%29%28x-6%29%3D0)
5. Apply the Distributive property. Then, you obtain:
![x^{4}-6x^{3}-16x^{2}+96x=0](https://tex.z-dn.net/?f=x%5E%7B4%7D-6x%5E%7B3%7D-16x%5E%7B2%7D%2B96x%3D0)
6. The polynomial of degree 4 and zeros −4, 0, 4 and 6 is:
![P(x)=x^{4}-6x^{3}-16x^{2}+96x](https://tex.z-dn.net/?f=P%28x%29%3Dx%5E%7B4%7D-6x%5E%7B3%7D-16x%5E%7B2%7D%2B96x)