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Math Help - remainder & factor theorem

  1. #1
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    remainder & factor theorem

    i understand the concepts and all the theorems for this topic, i am able to do some questions, but have difficulty with others, and also factorising cubic equations from scratch. thanks for helping will +reps for everyone, over a period of a few days(:

    1. Given that f(x)=x^2n - (p+1)x^2 +p where n and p are positive. Show that (x+1) is a factor of f(x) for all values of p. When p=4, find the value of n for which (x+2) is a factor of f(x) and, for this case, factorise f(x) completely.

    2. The expression 2x^3 - 3x^2 +ax +2 leaves a remainder of R when divided by 2x+1 and a remainder of 4R+5 when divided by x-3. Find the value of a.

    3. Factorise the expression 6x^3 - 7x^2 - x +2 fully. Hence find all the factors of 6(y+1)^3 - 7(y+1)^2 -y+1,
    solve the equation 3(2)^3p+1 + 2 = 7(2)^2p + 2^2p, giving the values of p to two decimal places where necessary.

    4. find the value of p for which (2+p)x + 6px^2 + p^2 -4p is divisible by x-1 but not x+1.
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  2. #2
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    Quote Originally Posted by swimmingskies View Post
    i understand the concepts and all the theorems for this topic, i am able to do some questions, but have difficulty with others,
    which part are you able to do and which part are you having difficulty with?
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  3. #3
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    Quote Originally Posted by swimmingskies View Post
    3. Factorise the expression 6x^3 - 7x^2 - x +2 fully. Hence find all the factors of 6(y+1)^3 - 7(y+1)^2 -y+1,
    solve the equation 3(2)^3p+1 + 2 = 7(2)^2p + 2^2p, giving the values of p to two decimal places where necessary.
    I'll start #3. Using the rational zeros theorem reveals that all of the roots are rational: -1/2, 2/3, and 1. So the factorization is
    6x^3 - 7x^2 - x + 2 = 6\left(x + \frac{1}{2}\right)\left(x - \frac{2}{3}\right)\left(x - 1\right).

    Now 6(y+1)^3 - 7(y+1)^2 - y + 1 = 6(y+1)^3 - 7(y+1)^2 - (y + 1) + 2, so
    6(y+1)^3 - 7(y+1)^2 - y + 1
    \begin{aligned}<br />
&= 6\left[(y + 1) + \frac{1}{2}\right]\left[(y + 1) - \frac{2}{3}\right]\left[(y + 1) - 1\right] \\<br />
&= 6y\left(y + \frac{3}{2}\right)\left(y + \frac{1}{3}\right)<br />
\end{aligned}


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    Last edited by yeongil; July 23rd 2009 at 05:50 AM.
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  4. #4
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    Quote Originally Posted by swimmingskies View Post
    2. The expression 2x^3 - 3x^2 +ax +2 leaves a remainder of R when divided by 2x+1 and a remainder of 4R+5 when divided by x-3. Find the value of a.
    Divide (2x^3 - 3x^2 +ax +2) \div (2x + 1) using long division and you get x^2 - 2x + \frac{a + 2}{2} as the quotient and 2 - \frac{a + 2}{2} = R as the remainder.

    You can also use long division when dividing (2x^3 - 3x^2 +ax +2) \div (x - 3), but using the remainder theorem will be quicker:
    2(3)^3 - 3(3)^2 +a(3) +2
    \begin{aligned}<br />
&= 2(27) - 3(9) + 3a + 2 \\<br />
&= 54 - 27 + 3a + 2 \\<br />
&= 3a + 29 = 4R + 5<br />
\end{aligned}

    Plug R = 2 - \frac{a + 2}{2} into the equation 3a + 29 = 4R + 5 and solve for a.


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  5. #5
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    Quote Originally Posted by swimmingskies View Post
    i understand the concepts and all the theorems for this topic, i am able to do some questions, but have difficulty with others, and also factorising cubic equations from scratch. thanks for helping will +reps for everyone, over a period of a few days(:

    1. Given that f(x)=x^2n - (p+1)x^2 +p where n and p are positive. Show that (x+1) is a factor of f(x) for all values of p. When p=4, find the value of n for which (x+2) is a factor of f(x) and, for this case, factorise f(x) completely.
    You titled this "remainder and factor theorem" so presumably you know that the remainder when polynomial P(x) is divided by x-a is P(a). In particular, x-a is a factor of P(x) if and only if P(a)= 0.
    f(1)= (-1)^{2n}- (p+1)(-1)^2+ p= 1- p- 1+ p= 0 that is enough to tell you that x-(-1)= x+ 1 is a factor. If p= 4, this is f(x)= x^{2n}- 5x^2+ 4. x+2 is a factor if and only if x= -2 satisfies the equation x^{2n}- 5x^2+ 4= 0. That is, (-2)^{2n}- 5(-2)^2+ 4= 4^n- 20+ 5= 4^n- 16= 0 or 4^n= 16. That should be easy to solve.

    2. The expression 2x^3 - 3x^2 +ax +2 leaves a remainder of R when divided by 2x+1 and a remainder of 4R+5 when divided by x-3. Find the value of a.
    2x+ 1= 2(x+ 1/2)= 2(x-(-1/2)) leaves remainder R so P(x) divided by x+1/2 would leave remainder 2R. P(-1/2)= -2/8- 3/4- (1/2) a+ 2= (1/2)a- (3/2)= R or, multiplying by 4, 2a- 6= 4R. Similarly, since P(x) divided by x-3 leaves remainder 4R+ 2, we have P(3)= 2(3^3)- 3(3^2)+ 3a+ 2= 3a+ 29= 4R+5 or 3a- 5= 4R. 4R= 2a- 6= a- 5 and now you can solve for a.

    3. Factorise the expression 6x^3 - 7x^2 - x +2 fully. Hence find all the factors of 6(y+1)^3 - 7(y+1)^2 -y+1,
    solve the equation 3(2)^3p+1 + 2 = 7(2)^2p + 2^2p, giving the values of p to two decimal places where necessary.
    Once more, it is easy to see that [tex]6(1^3)- 7(1^2)- (1)+ 2= 6- 7- 1+ 2= 0 so x-1 is a factor. Factoring out x- 1, it is easy to see that the leading term of the other factor must be 6x^2 and the constant term
    must be -2. That leaves only the coefficient of x to be determined: [tex](x-1)(6x^2+ ?x- 2)= 6x^3- (6- ?)x^2- (?+2)x+ 2. So we must have 6- ?= 7 and ?+ 2= 1 so both equations give ?= -1.
    6x^3- 7x^2- x+ 2= (x- 1)(x^2- x- 2). You can use the quadratic formula to find the factors of x^2- x- 2.

    4. find the value of p for which (2+p)x + 6px^2 + p^2 -4p is divisible by x-1 but not x+1.
    If it is divisible by x- 1 than (2+p)(1)+ 5p(1)^2+ p^2- 4p= 0so p^2+ 3p+ 3= (p+2)(p+1). p is either -2 or -1. If p were equal to -1, x= -1 would also make that polynomial 0 so x+ 1 would be a factor. Since it is not, p must be -2.
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