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Math Help - Issue with a factoring problem

  1. #1
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    Issue with a factoring problem

    Factor completely.

    all variables have an exponent which follows it

    20x4 - 58x5 + 42x4

    it would seem to me to factor a 2x to the fourth with a result of:

    2x4(10x2 - 29x + 21)

    again, all variables are followed by the exponent

    Any help in understanding what the problem is would be of great appreciation. To me everything appears fine. If you distribute 2x to the fourth across, you end with the original expression.

    However my homework web site tells me it's incorrect so I can only trust that they are right.

    thanks again.

    :Edited for clarity:
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  2. #2
    Super Member Quacky's Avatar
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    Re: Issue with a factoring problem

    Quote Originally Posted by helpwanted View Post
    Factor completely.

    all variables have an exponent which follows it

    20x4 - 58x5 + 42x4

    it would seem to me to factor a 2x to the fourth with a result of:

    2x4(10x2 - 29x + 21)

    again, all variables are followed by the exponent

    Any help in understanding what the problem is would be of great appreciation. To me everything appears fine. If you distribute 2x to the fourth across, you end with the original expression.

    However this web site tells me it's incorrect so I can only trust that they are right.

    thanks again.
    You started with 20x^4-58x^5+42x^4

    You then stated that this is equivalent to:

    2x^4(10x^2 - 29x + 21)

    If we multiply out your expression, we get:

    2x^4(10x^2)+2x^4(-29x)+2x^4(21)

    =20x^6-58x^5+42x^4

    Which is not what you had originally.

    Did you intend to write the problem statement as:

    20x^6-58x^5+42x^4?
    In which case, your answer is correct, but not complete.

    Notice the quadratic in the brackets:

    10x^2 - 29x + 21
    Taking the discriminant:

    29^2-4(10)(21)
    =1

    The discriminant is a perfect square, therefore we conclude that this quadratic, surprisingly, factors.
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  3. #3
    Junior Member mathbyte's Avatar
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    Re: Issue with a factoring problem

    All I will say is to double check your answer by multiplying out your factored answer.
    When multiplied out, one of the terms doesn't match.
    Also, remember you can collect like terms before factoring.
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  4. #4
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    Re: Issue with a factoring problem

    Yes you are correct, I miss typed it, it is

    20x^6-58x^5+42x^4

    but I don't see the factors of:

    2x^4(10x^2 - 29x + 21)

    I'm now looking for factors of (10x^2 - 29x + 21).. and I'm failing to see the connection
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  5. #5
    Super Member Quacky's Avatar
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    Re: Issue with a factoring problem

    Quote Originally Posted by helpwanted View Post
    Yes you are correct, I miss typed it, it is

    20x^6-58x^5+42x^4

    but I don't see the factors of:

    2x^4(10x^2 - 29x + 21)

    I'm now looking for factors of (10x^2 - 29x + 21).. and I'm failing to see the connection
    Factors of 10x^2-29x+21 are factors of 2x^4(10x^2 - 29x + 21) and therefore, they are factors of 20x^6-58x^5+42x^4, which are what you have been asked to identify.

    It's like saying, here's the number 36 and 36=9\times{4} therefore the factors of 36 are 9 and 4. That's correct, and true, but it isn't complete. The factors of 9 and 4 are also factors of 36, amongst others.

    For this quadratic, I'd suggest using the formula to find the roots, and deduce the factors from there, rather than use the "guess and check" method. I told you previously that the discriminant is 1, which should save you some time.
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  6. #6
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    Re: Issue with a factoring problem

    Well thank you for your time Quacky I really appreciate it. I edited my first post because it made it seem that mathhelpforum.com was telling me my answer was wrong, and what I meant to say was that my homework web site was, and aparently is correct at that. I however, do not see any way to further factor this problem more than I already have =/, I suppose that's why I'm always a C student in math.

    Thanks again for your help,


    -Park
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  7. #7
    Super Member Quacky's Avatar
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    Re: Issue with a factoring problem

    I think it's more likely that you're a C student because you accept defeat too quickly!

    10x^2-29x+21

    Take f(x)=10x^2-29x+21

    Now then, when f(x)=0,

    10x^2-29x+21=0

    In this situation, \frac{-b\pm\sqrt{b^2-4ac}}{2a}=0

    Now, you could go through and do the substitution - you'd end up with x=\frac{29\pm{1}}{20}

    Either:

    x=\frac{30}{20} so x=\frac{3}{2}

    or x=\frac{28}{20} so x=\frac{7}{5}

    Now, if a quadratic factors, then we have something of the form (a'x+b')(cx+d)=0 (the ' are just there to prevent repetition of a and b)

    Now we know that when (a'x+b')(cx+d)=0, x=\frac{3}{2} firstly. This means that when x=\frac{3}{2}, one of those brackets is equal to 0.

    I.e. (a'\frac{3}{2}+b')=0

    You should soon be able to deduce that a'=2 and b'=-3 is one valid solution. Once you have that, working out (cx+d) should be even more straightforward because you know that (2x-3)(cx+d)=10x^2-29x+21 so by observation, c=5 and d=-7 to give us our factors:

    10x^2-29x+21=(2x-3)(5x-7)

    Alternatively, you could use the guess-and-check method:
    Factors of 10:
    1,10
    2,5

    Factors of 21:
    1,21
    3,7

    You need to take one pair of each numbers and multiply them together in such a way that the multiplied numbers:
    -Add up/subtract to 29
    -Multiply to 21

    From there, you can just adjust the signs.
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  8. #8
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    Re: Issue with a factoring problem

    Ayee... I see now [light bulb!], thank for such a thorough response. The upper half of the response was above my current level, however the bottom portion makes perfect sense. For some reason when they said factor completely I was stuck with looking for the greatest common factor, and seeing none, I stopped, completely forgetting to factor the trinomial.

    Your response and time is very much appreciated.
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  9. #9
    Super Member Quacky's Avatar
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    Re: Issue with a factoring problem

    Quote Originally Posted by helpwanted View Post
    Ayee... I see now [light bulb!], thank for such a thorough response. The upper half of the response was above my current level, however the bottom portion makes perfect sense. For some reason when they said factor completely I was stuck with looking for the greatest common factor, and seeing none, I stopped, completely forgetting to factor the trinomial.

    Your response and time is very much appreciated.
    That's a fairly common trail of thought; don't be disheartened by it. In fact, it's a very deceiving question: 10x^2-29x+21 isn't something that looks as though it factors - I wouldn't have spotted the factors if not for the fact that your homework checker caught it out.
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  10. #10
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    Re: Issue with a factoring problem

    Ok; one a little easier (so you feel better!); factor completely:
    5x^6 - 25x^5 + 30x^4
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  11. #11
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    Re: Issue with a factoring problem

    I have another factoring issue. 32 has an exponent of 2.

    [PHP].08r - .04 = -.32r^2[/PHP]

    multiply by 100 to get rid of the decimals:

    [PHP]8r - 4 = 32r^2[/PHP]

    set the equation equal to zero:

    [PHP]32r^2 + 8r - 4 = 0[/PHP]

    factor the greatest common factor:

    [PHP]
    4(8r^2 + 2 - 1) = 0

    //--->or

    -4(-8r^2 - 2 +1) = 0[/PHP]

    The above does not factor for me no matter how I try =/. The question is expecting two answers for R. A smaller number, and a Larger number. Thanks again.
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  12. #12
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    Re: Issue with a factoring problem

    Hello, helpwanted!

    \text{Factor completely: }\:20x^6 - 58x^5 + 42x^4

    Factor out 2x^4\!:\quad 2x^4(10x^2 - 29x + 21)

    The quadratic expression does factor.
    Do you mean you have never been taught how to factor a trinomial?

    Here's a method you might like . . . or maybe not.


    Multiply the leading coefficient (10) by the constant (21): (10)(21) \,=\,210

    Factor 210 into two parts whose sum is the middle coefficient (29).

    How do we find that pair?
    Divide by 210 by 1, 2, 3, etc., keeping the ones that "come out even",
    . . and look for the pair whose sum is 29.

    . . \begin{array}{cccc}\text{Divide} & \text{Factors} & \text{Sum} \\ \hline 210 \div 1 & 1\cdot210 & 211 \\ 210 \div 2 & 2\cdot105 & 107 \\ 210 \div 3 & 3\cdot70 & 73 \\ 210 \div 4 & --- & -- \\ 210 \div 5 & 5\cdot42 & 47 \\ 210 \div 6 & 6\cdot35 & 41 \\ 210 \div 7 & 7\cdot30 & 37 \\ 210 \div 8 & --- & -- \\ 210 \div 9 & --- & -- \\ 210 \div 10 & 10\cdot21 & 31 \\ 210 \div 11 & --- & -- \\ 210 \div 12 & --- & -- \\ 210 \div 13 & --- & -- \\ 210 \div 14 & 14\cdot15 & 29 \\ && \uparrow \\ && \text{There!} \end{array}


    Replace the middle term (-29x) with -14x and -15x.

    We have: . 10x^2 - 14x - 15x + 21

    Factor by grouping:
    . . . . . . . 2x(5x - 7) - 3(5x - 7)

    Factor: . (5x-7)(2x-3)

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  13. #13
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    Re: Issue with a factoring problem

    Thanks Soroban. I find my mind likes to work mathmatical problems out through procedure, and for some reason while factoring I have a really hard time maintaining a procedure that brings me to the correct outcome. I will give your example a shot in a few problems.

    Thanks again.
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  14. #14
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    Re: Issue with a factoring problem

    Quote Originally Posted by helpwanted View Post
    4(8r^2 + 2 - 1) = 0
    That should be 8r^2 + 2r - 1 = 0

    Use the quadratic formula to solve:
    r = [-2 +- SQRT(2^2 - 4(8)(-1)] / [2(8)]

    You seem unaware of the basics; attending math classes?
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