how do u solve (2-x+3x^2)(3+2x-x^2+2x^4)

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how do u solve (2-x+3x^2)(3+2x-x^2+2x^4)

Mathematics
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whats your definition of solve?
to me; solve would be to distribute each term in the left thru the paranethesis and then comcine like terms
sorry Find the first three terms of the product in ascending powers (2-x+3x^2)(3+2x-x^2+2x^4)

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Other answers:

ascending eh...from little power to biggest...
yes
6 +4x -3x+9x^2-2x^2-2x^2...if I see it right :) 6 +x +5x^2
and I think i see others....hold that
3 +2x -x^2 * 2 -x +3x^2 -------------- 9x^2+0+0 -3x-2x^2+0 6+4x-2x^2 ------------------- 6 +x +5x^2.....yeah, that should be it
thanks you
can u help me with one more
i can try
and this time let me do it and you do it also and then we can compare ou answers to see if i got it
k
(1-2x^2-x^4)(4+x^2-2x^4)
i got mine :)
i got mine
so i got 4+9x^2+5x^6
4 - 6x^2 -8x^4 ??
is it right
the lowest powers are x^0; x^2 and x^4
4 - 7x^2 -8x^4 ...had a typo :) 4x^0 = 4 x^2 -8x^2 = 7x^2 -2x^4 -2x^2 -4x^4 = -8x^4
0k - got it
thanks

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