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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

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AT FIRST. do u know what a degree of polynomial means.
\[3x^3+(3x^2/2)−(3x/2)+(3/2)\] is this correct?
for 1.?

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Other answers:

simplify that.
Is my answer correct?
how about for no. 2 could you help me?
can u simplify this plz
FOR NUMBER 2) do u know factor theorem
could I make \[(x-1/2) -------> (2x-1)\] ????????
NOW do u know factor theorem? and did u get answer for 1)
am I correct now?
I'll do it again :)
u just made the sign mistake
could I make (x-1/2) into (2x-1) because my teacher said there should be no fractions?
is the correct answer
could I make (x-1/2) into (2x-1) ??????
NOPE because|dw:1343208355724:dw|
but if |dw:1343208386811:dw|
that's what I mean for the second part
isn't that it is zero so that it is possible?
IF it is equal to zero then yes u can because |dw:1343208534245:dw|
so what?
I got the correct answer :)
and what is that
it mean that there zeroes are both 1/2
so you can use either of the two?
yeah. BUT only in that case. U cannot use this in simplification.
yup :)
do u know factor theorem.
wait I am trying to solve 1 with (2x-1) :)
I got the same answer :)
I know factor theorem. If the remainder when dividing P(x) by D(x) is zero, then D(x) is a factor of P(x). P(x) is any polynomial D(x) is the divisor being a factor means it is included in multiplying.
is that correct? could you improve my definition of factor ?
No u r defination is correct. And can u use this to solve this problem
how I am confused with the question?
I am thinking of distributing -3 to x
but I can't explain why it is correct :)
your 2 number question simply means that if u divide P(x) by (x+3) then remainder is zero. |dw:1343209619170:dw|
ohh, I think it also has remainder theorem? :)
now can u
I added =0 so that the remainder will be equal to zero making (x+3) a product ?
solve for k
am I correct with my reason?
@sauravshakya thank you very much for your help :)
U r WELCOME. AND best of luck

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