1.) Find a polynomial of degree that has zeros 1, -1 and 1/2 and whose leading coefficient is 3. 2.) Find the value of k such that x+3 is a product of P(x) = x^2 - 10kx + 6

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1.) Find a polynomial of degree that has zeros 1, -1 and 1/2 and whose leading coefficient is 3. 2.) Find the value of k such that x+3 is a product of P(x) = x^2 - 10kx + 6

Mathematics
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My answer for no. 1 is \[3x^3 -( 3x^2/2) - 3x + (3/2)\] but my teacher said there should be no fractions
then multiply all the terms by 2(the denominator) to get rid of the fraction
but the leading coefficient will not be 3???

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Other answers:

@saifoo.khan
could you help me?
|dw:1343106002554:dw|
What does it mean by leading coefficient?
http://4.bp.blogspot.com/_lvMAEEKgEx8/TM6ebuxxeYI/AAAAAAAAAhA/PMyxtcHq5hI/s320/a.gif
isn't that your answer does not have a leading coefficient of 3 ?
you did a mistake in multiplying.
(x-1)(x+1)(x-1/2) = 0 try again.
I'll do it again :)
Sure.
I did (x-1)(x+1)(2x-1) = 0 is it the same as (x-1)(x+1)(x-1/2) = 0 ??
yup.
I got\[x^3+( x^2/2) -(x/2)+(1/2)\] I checked it twice :)
@saifoo.khan is this correct?
Now this is correct!! :D
NOw get rid of 2 in the denominators, multiply the whole expression by 2.
\[2x^3 + x^2 - x + 1\]
How will I make it into 3?
Are decimals allowed?
he said there is no fractions
SInce your teacher said NO FRACTIONS.
:D
that's why I am confused :) or maybe I misheard him :)
Possibly..
We can multiply the equation by 1.5 to make the leading coef as 3. :D
maybe he is referring to (2x-1) and (x-1/2)
2x-1 and x-1/2 is same.
are* same.
so my answer I think is \[3x^3 + (3x^2/2) - (3x/2) + (3/2)\] that's the only expression I can have to make the leading coefficient equal to 3
is that correct?
3x^3 + 1.5x^2 - 1.5x + 1.5 NO FRACTIONS!!! :D
lol, I'll just include that to my answer :)
how about for the second one ?
HAHA. http://funny-pictures-blog.com/wp-content/uploads/funny-pictures/Like-a-boss.jpg
That question looks like a challenge!!
it should be P(x) = x^2 - 10kx + 6
it had a typo :)
Find the value of k such that x+3 is a product of P(x) = x^2 - 10kx + 6 (-3)^2 - 10k(-3) + 6 = 0 9 + 30 k + 6 = 0
I got k = 1/2 but why did you made x into -3?
Because since it's a product of it, the remainder is 0.
So we insert the factor and set the equation = 0.
Thanks, I understand it :)
Great. Welcome.

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