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