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January 28, 2015

January 28, 2015

Posted by **allison--Please help!!** on Thursday, December 27, 2012 at 5:08pm.

I put: x^2(x-3)+16(x=3)

(x-3)(x^2+16)

For zeros:

x-3=0

x=0

**My teacher stated check the equation solution again. What is the value for x and hence what is the zero for the polynomial based on the first factor?

**Can someone help me?

- calculus-can someone please help me with this ques -
**drwls**, Thursday, December 27, 2012 at 6:09pmYou factored the cubic polynomial correctly as

(x-3)(x^2 + 16)

The first term tells the function has a zero at x = 3.

The other roots are imaginary. x = +/- 4i

- calculus-can someone please help me with this ques -
**allison--Please help!!**, Thursday, December 27, 2012 at 6:56pmyes, I did put:

x^2+16=0

x^2= -16

x= + or -4i

But she wrote that on the first part that I submitted and I don't really know what it is that she wants me to do.

**So are you saying that the value for x is 3, but then what is the zero for the polynomial based on the first factor??

- calculus-can someone please help me with this ques -
**Damon**, Thursday, December 27, 2012 at 7:33pmx-3 = 0

means x = 0

so the polynomial is zero at x = 3

also at - 4 i and at + 4 i

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