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Thread: Several algebra questions regarding factorisation

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Several algebra questions regarding factorisation

    A very good day to all.
    I have several questions all relating to algebra, and I will be very thankful if you can help me solve them. Please also show your workings clearly so that I may understand wholly. I am sorry if these questions take a lot a time.

    Factorise the following:

    1.$\displaystyle (x + y)(a + b) - (y + z)(a + b)$


    2. $\displaystyle (2x + y)^2 -3(2x + y)$


    3. $\displaystyle 5(m - 2n) - (m -2n)^2$


    4. $\displaystyle p^2 -9q^2 +6qr -r^2$


    5. $\displaystyle 4a^2 -b^2 -2bc -c^2$
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by PythagorasNeophyte View Post
    A very good day to all.
    I have several questions all relating to algebra, and I will be very thankful if you can help me solve them. Please also show your workings clearly so that I may understand wholly. I am sorry if these questions take a lot a time.

    Factorise the following:

    1.$\displaystyle (x + y)(a + b) - (y + z)(a + b)$
    Observe that $\displaystyle (a+b)$ is the common factor. So $\displaystyle (x+y)(a+b)-(y+z)(a+b)=(a+b)(\ldots)$

    2. $\displaystyle (2x + y)^2 -3(2x + y)$
    Observe that $\displaystyle (2x+y)$ is the common factor. So $\displaystyle (2x+y)^2-3(2x+y)=(2x+y)(\ldots)$

    3. $\displaystyle 5(m - 2n) - (m -2n)^2$
    Observe that $\displaystyle (m-2n)$ is the common factor. So $\displaystyle 5(m-2n)-(m-2n)^2=(m-2n)(\ldots)$

    4. $\displaystyle p^2 -9q^2 +6qr -r^2$
    This one is slightly more complicated. First observe that $\displaystyle p^2-9q^2+6rq-r^2=p^2-(9q^2-6qr+r^2)$.

    Now, $\displaystyle 9q^2-6qr+r^2=(3q-r)^2$. So now we are left with $\displaystyle p^2-(3q-r)^2$, which can be factored using the difference of squares formula.


    5. $\displaystyle 4a^2 -b^2 -2bc -c^2$
    This is similar to #4. Observe that $\displaystyle 4a^2-b^2-2bc-c^2=(2a)^2-(b^2+2bc+c^2)$.

    Now, $\displaystyle b^2+2bc+c^2=(b+c)^2$. So now we are left with $\displaystyle (2a)^2-(b+c)^2$, which can be factored using the difference of squares formula.

    Does this make sense? Can you try these problems now?
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  3. #3
    Junior Member PythagorasNeophyte's Avatar
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    Thank you so much for your help Chris!

    However, I still could not work out questions 1, 2 and 3 myself. I know the common factors but I am not sure about the other terms.

    Also, I would appreciate if you could verify my answers for question 4 and 5.
    The answer for question 4 might be---

    $\displaystyle p^2 -(3q -r)^2 $
    $\displaystyle =[p^2 +(3q-r)][p^2 -(3q -r)]$
    $\displaystyle =(p^2 +3q -r)(p^2 -3q +r)$

    Whereas for question 5---

    $\displaystyle (2a)^2 -(b+c)^2$
    $\displaystyle =[2a +(b+c)][2a -(b +c)$
    $\displaystyle = (2a +b +c)(2a -b -c)$

    Is my answers and workings correct?
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by PythagorasNeophyte View Post
    However, I still could not work out questions 1, 2 and 3 myself. I know the common factors but I am not sure about the other terms.
    1
    $\displaystyle (a+b)(x+y-y-z)=(a+b)(x-z)$

    2
    $\displaystyle (2x+y)^2-3(2x+y)=(2x+y)(2x+y)-3(2x+y)=(2x+y)(2x+y-3)$

    3
    same as second
    $\displaystyle (m-2n)(5-m+2n)$
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