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Mathematics 7 Online
OpenStudy (anonymous):

* I know that I need to use the factor theorem for A, but I'm not sure how to do it. * Someone told me that B would be 3x-1, C would be (x+3)(x-2)(3x-1) and that D would be -3,2 and 1/3 but I'm not sure how they got those answers. Can someone please help? a) verify the given factors of the function f, b) find the remaining factors of f, c) use your results to write the complete factorization of f, d) list all real zeros of f, and e) confirm your results by using a graphing utility to graph the function. Function: f(x)=3x^3+2x^2-19x+6 Factors: (x+3), (x-2)

OpenStudy (anonymous):

are you given the two factor to start with?

OpenStudy (anonymous):

yes the very last part "Function: f(x)=3x^3+2x^2-19x+6 Factors: (x+3), (x-2)" is the given function and given factors

OpenStudy (anonymous):

ok then you know it has to be \[(x+3)(x-2)(ax+b)=3x^3+2x^2-19x+6\]

OpenStudy (anonymous):

is that part ok?

OpenStudy (anonymous):

i think so, what would i do with the 3x^3+2x^2-19x+6 part though?

OpenStudy (anonymous):

btw what we're doing right now, is that the factor theorem for part A?

OpenStudy (anonymous):

you are given two factors, which means you know that \[ 3x^3+2x^2-19x+6\] factors and the factor are \(x+3\) and \(x-2\) making \[ 3x^3+2x^2-19x+6=(x+3)(x-2)(something)\] your job here is to find the something

OpenStudy (anonymous):

the polynomial \(3x^3+2x^2-19x+6\) has degree 3 and \((x+3)(x-2)\) is a polynomial of degree 2 that mean the "something" is a polynomial of degree 1, i.e a line, in the form \(ax+b\) and your job here is to find \(a\) and \(b\)

OpenStudy (anonymous):

that is why i wrote \[(x+3)(x-2)(ax+b)=3x^3+2x^2-19x+6\]

OpenStudy (anonymous):

it should be pretty clear what \(a\) and \(b\) are because on the left you have \((x+3)(x-2)\) starts with \(x^2\) and on the right you have \(3x^3\) which tells you \(a=3\) which makes it \[(x+3)(x-2)(3x+b)=3x^3+2x^2-19x+6\]

OpenStudy (anonymous):

as for \(b\) on the left you have \(+3\times -2=-6\) and on the right you have \(+6\) which tells you \(b=-1\) making it \[(x3)(x-2)(3x-1)=3x^3+2x^2-19x+6\]

OpenStudy (anonymous):

typo there, i means \[(x+3)(x-2)(3x-1)=3x^3+2x^2-19x+6\]

OpenStudy (anonymous):

thank you for taking the time to explain this!! so was all of this that you just did part A?

OpenStudy (anonymous):

or if it was A, B, and C like I think it was, how would I do part D ( list all real zeros of f)?

OpenStudy (marigirl):

satellite73: u wrote: "as for b on the left you have +3×−2=−6 and on the right you have +6 which tells you b=−1 making it (x3)(x−2)(3x−1)=3x3+2x2−19x+6" when u wrote: "and on the right you have +6 ... are u referring to the constant term of the equation?

OpenStudy (marigirl):

@satellite73

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