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    How do you factor: 2x^2n + 7x^n – 15 ?

    I'm confused because I've never worked it out when the equation has powers (^2n...) :/

    Thanks!!!
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    Hey Rosa

    You know how you factorise 2x2 + 7x - 15? You can literally do the same thing, and at the end just replace your xs with xns
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    (Original post by Student403)
    Hey Rosa

    You know how you factorise 2x2 + 7x - 15? You can literally do the same thing, and at the end just replace your xs with xns
    Okay, I've got it Pretty simple, now that I understand
    Thank you!
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    (Original post by RosaA)
    Okay, I've got it Pretty simple, now that I understand
    Thank you!
    My pleasure and thanks for the tag

    I hope you understand the reason why btw!
    Spoiler:
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    Because 2x2n = 2(xn)2 and so on

    You could do the same with a quadratic involving anything like that instead of an x.. Even a function like 2cos2θ +7cosθ - 15
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    (Original post by RosaA)
    How do you factor: 2x^2n + 7x^n – 15 ?

    I'm confused because I've never worked it out when the equation has powers (^2n...) :/

    Thanks!!!
    In addition to 403's excellent answer: This is called a disguised quadratic (in this case, it's a quadratic in x^n). What this fancy shmoozy wording means is that if you make a substitution (I always make this explicit substitution when working out problems like this) u=x^n then you have a quadratic in u or a quadratic in x^n written as 2u^2 + 7u - 15= 0 .

    I'd make the substitution, re-write everything in terms of u and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to x by back-substitution u=x^n.
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    Oh, and moved to maths.
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    (Original post by RosaA)
    How do you factor: 2x^2n + 7x^n – 15 ?

    I'm confused because I've never worked it out when the equation has powers (^2n...) :/

    Thanks!!!
    I'll show you an example - suppose the case where n = 2, and we have the equation \displaystyle 2x^{4}+7x^{2}-15 = 0.

    Now, we can factorise this the regular way (i.e. the way we do quadratic equations as 403 mentioned)

    \displaystyle 2x^{4}+ 10x^{2} - 3x^{2} -15= 0

    \displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0

    Now, it seems that we have two equations from which we can obtain factors:

    \displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0 \Longleftrightarrow 2x^{2}-3 = 0 \ \text{or} \ x^{2} +5 = 0

    We can now evaluate these two equations to obtain our roots, so with the first one,

    \displaystyle x^{2} = \frac{3}{2} \Longleftrightarrow x=\pm \sqrt{\left ( \frac{3}{2} \right )}, as taking a square root gives us two solutions.

    However, for the second equation, note that it leads to \displaystyle x^{2} = -5 and at GCSE level, you can't take the square root of negative numbers so we have to eliminate this solution \displaystyle \Rightarrow x^{2} \neq -5.

    We only have two solutions in this case: \displaystyle \therefore x=\pm \sqrt{\left ( \frac{3}{2} \right )}

    I hope this example helped!
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    (Original post by Zacken)
    In addition to 403's excellent answer: This is called a disguised quadratic (in this case, it's a quadratic in x^n). What this fancy shmoozy wording means is that if you make a substitution (I always make this explicit substitution when working out problems like this) u=x^n then you have a quadratic in u or a quadratic in x^n written as 2u^2 + 7u - 15= 0 .

    I'd make the substitution, re-write everything in terms of u and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to x by back-substitution u=x^n.
    PRSOM
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    (Original post by aymanzayedmannan)
    I'll show you an example - suppose the case where n = 2, and we have the equation \displaystyle 2x^{4}+7x^{2}-15 = 0.

    Now, we can factorise this the regular way (i.e. the way we do quadratic equations as 403 mentioned)

    \displaystyle 2x^{4}+ 10x^{2} - 3x^{2} -15= 0

    \displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0

    Now, it seems that we have two equations from which we can obtain factors:

    \displaystyle \left ( 2x^{2}-3 \right )\left ( x^{2} +5 \right ) = 0 \Longleftrightarrow 2x^{2}-3 = 0 \ \text{or} \ x^{2} +5 = 0

    We can now evaluate these two equations to obtain our roots, so with the first one,

    \displaystyle x^{2} = \frac{3}{2} \Longleftrightarrow x=\pm \sqrt{\left ( \frac{3}{2} \right )}, as taking a square root gives us two solutions.

    However, for the second equation, note that it leads to \displaystyle x^{2} = -5 and at GCSE level, you can't take the square root of negative numbers so we have to eliminate this solution \displaystyle \Rightarrow x^{2} \neq -5.

    We only have two solutions in this case: \displaystyle \therefore x=\pm \sqrt{\left ( \frac{3}{2} \right )}

    I hope this example helped!
    Yes, it was very helpful indeed!! Thank you so much !!
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    (Original post by Zacken)
    In addition to 403's excellent answer: This is called a disguised quadratic (in this case, it's a quadratic in x^n). What this fancy shmoozy wording means is that if you make a substitution (I always make this explicit substitution when working out problems like this) u=x^n then you have a quadratic in u or a quadratic in x^n written as 2u^2 + 7u - 15= 0 .

    I'd make the substitution, re-write everything in terms of u and then do my cool quadratic work with awesome quadratic skillz and then convert everything back to x by back-substitution u=x^n.
    I need your brain -literally Thanks so much!!
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    (Original post by Student403)
    My pleasure and thanks for the tag

    I hope you understand the reason why btw!
    Spoiler:
    Show
    Because 2x2n = 2(xn)2 and so on

    You could do the same with a quadratic involving anything like that instead of an x.. Even a function like 2cos2θ +7cosθ - 15
    No it's cool, no need to thank me -I owe you .

    Yup, I understand it now
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    (Original post by RosaA)
    No it's cool, no need to thank me -I owe you .

    Yup, I understand it now
    Love it when people say that :awesome:
 
 
 
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