# Thread: remainder & factor theorem

1. ## 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.

2. Originally Posted by swimmingskies
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?

3. Originally Posted by swimmingskies
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
$\displaystyle 6x^3 - 7x^2 - x + 2 = 6\left(x + \frac{1}{2}\right)\left(x - \frac{2}{3}\right)\left(x - 1\right)$.

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

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4. Originally Posted by swimmingskies
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 $\displaystyle (2x^3 - 3x^2 +ax +2) \div (2x + 1)$ using long division and you get $\displaystyle x^2 - 2x + \frac{a + 2}{2}$ as the quotient and $\displaystyle 2 - \frac{a + 2}{2} = R$ as the remainder.

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

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

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5. Originally Posted by swimmingskies
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.
$\displaystyle f(1)= (-1)^{2n}- (p+1)(-1)^2+ p= 1- p- 1+ p= 0$ that is enough to tell you that x-(-1)= x+ 1 is a factor. If p= 4, this is $\displaystyle f(x)= x^{2n}- 5x^2+ 4$. x+2 is a factor if and only if x= -2 satisfies the equation $\displaystyle x^{2n}- 5x^2+ 4= 0$. That is, $\displaystyle (-2)^{2n}- 5(-2)^2+ 4= 4^n- 20+ 5= 4^n- 16= 0$ or $\displaystyle 4^n= 16$. 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 $\displaystyle 6x^2$ 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.
$\displaystyle 6x^3- 7x^2- x+ 2= (x- 1)(x^2- x- 2)$. You can use the quadratic formula to find the factors of $\displaystyle x^2- x- 2$.

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 $\displaystyle (2+p)(1)+ 5p(1)^2+ p^2- 4p= 0$so $\displaystyle p^2+ 3p+ 3= (p+2)(p+1)$. 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.