Factor the trinomial c^3-2c^2-48c?? HELP!
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Factor the trinomial c^3-2c^2-48c?? HELP!

[From: ] [author: ] [Date: 12-02-21] [Hit: ]
then it becomes easy but with this one its dead easy, notice there is no numerical term so everything has c as a factor.now its just factorising a straight forward quadratic.Two numbers that are two apart and multiply to give 48, well we know from our times table that thats 6*8, now its -2 so the 8 must be the -ve one.......
I have tried to factor this and I do not understand how to factor a cube????

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c(c-8)(c+6)

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Normally I have to do a bit of guess work to find the first factor, then it becomes easy but with this one it's dead easy, notice there is no numerical term so everything has c as a factor.

c(c^2 - 2c - 48)

now it's just factorising a straight forward quadratic.

Two numbers that are two apart and multiply to give 48, well we know from our times table that that's 6*8, now it's -2 so the 8 must be the -ve one. They must be opposite signs as the numerical term is negative.

(c-8)(c+6) = c^2 + 6c -8c - 48 = c^2 -2c - 48 - so yes that works.

Your factors then are c(c-8)(c+6)
hope this helps

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c(c^2-2c-48)
c(c^2+6c-8c-48)
c[c(c+6)-8(c+6)]
c(c+6)(c-8)

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c³- 2c² - 48c = c(c² - 2c - 48)

=> c(c - 8)(c + 6)

:)>
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