# Give a possible formula for this 4th-degree polynomial...

• Sep 8th 2010, 07:24 PM
melliep
Give a possible formula for this 4th-degree polynomial...
Assume the graph of a 4th-degree polynomial crosses the x-axis at exactly -2, 2, and 5. Let k be the leading coefficient with k > 0.

Give a possible formula for this 4th-degree polynomial.
• Sep 8th 2010, 07:29 PM
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Quote:

Originally Posted by melliep
Assume the graph of a 4th-degree polynomial crosses the x-axis at exactly -2, 2, and 5. Let k be the leading coefficient with k > 0.

\$\displaystyle P(x)=k(x+2)^a(x-2)^b(x-5)^c\$

let (a,b,c)=(1,1,2) or (1,2,1) or (2,1,1)
• Sep 8th 2010, 08:05 PM
melliep
I tried that, it was incorrect =(
Nothing I do is working.
• Sep 8th 2010, 08:10 PM
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Quote:

Originally Posted by melliep
I tried that, it was incorrect =(
Nothing I do is working.

Was it rejected by an automatic computer checker? Was it marked incorrect by a teacher? Did it not match an answer key? If you were expected to expand, are you sure you did it without making mistakes?

Because it certainly looks correct from where I'm standing..
• Sep 8th 2010, 08:27 PM
melliep
it was rejected on online homework.
haha, yes I input it correctly, I'm sure.
I'm not sure why it's consistently wrong. But thank you very much.
• Sep 8th 2010, 08:32 PM
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Quote:

Originally Posted by melliep
it was rejected on online homework.
haha, yes I input it correctly, I'm sure.
I'm not sure why it's consistently wrong. But thank you very much.