# Thread: Remainder and Factor Theorem HELP

1. ## Remainder and Factor Theorem HELP

well i missed a week off school, and i am completely in the dark on my homework, if someone could please tell me what to do it would be GREATLY appreciated
1. The function f is given by f(x) = 2x 3 + x 2 – 13x + 6.
(a) Find f(–2), f(½) and f(2).
(i) Write down the remainder when f(x) is divided by x + 2.
(ii) Explain why 2x – 1 is a factor of f(x).
(b) Factorise f(x) and hence solve f(x) = 0.
(c) Sketch the curve y = f(x).

2. Factorise p(x) = 13x - 6 - 4x 3. HINT you may have to divide for this one!
Solve p(x) = 0.
Sketch the graph of y = 13x - 6 - 4x 3.

3. (a) Use the remainder theorem to find the remainder when x 3 – 2x 2 + 8x – 3 is divided by x – 5.
(b) When x 3 – 4x 2 + ax + 6 is divided by x – 3 the remainder is 18.
Find the value of the constant a.

4. When x 3 + px 2 + qx + 1 is divided by x – 2 the remainder is 9 and when divided by x + 3
the remainder is 19. Find the values of the constants p and q.

5. Find the value of the constant k given that x + 4 is a factor of f(x), where
f(x) = 3x 3 + 10x 2 – kx - 4.
Hence factorise f(x) and solve f(x) = 0.

2. f(x) is the given function. I'll try to help you with the first one, and hopefully you'll understand how to do it from there.
a.)
$\displaystyle f(x) = 2x^3 + x^2 -13x + 6$
$\displaystyle f(-2) = 2(-2)^3 +(-2)^2 -13(-2) +6$ Plug in -2 whenever you see x

$\displaystyle f(-2) = -16 +4 +26+6 = 20$, might be off on the arithmetic
Do the same for the other values, just plug in the given value whenever you see a x and simplify.
i.)
Polynomial Long Division

Here is a website that shows you how to do polynomial long division.
ii.)
If there is no remainder, than the divisor is a factor of the entire polynomial. (e.g. 24/3 = 8, no remainder, so 3 is a factor)
b.)
To factor the polynomial, you divide it by its factors, and you are already given one. Divide first by $\displaystyle (2x-1)$, and you will get a quadratic equation of the form $\displaystyle ax^2 + bx + c$, and that should be simply to factor.

Once you have it all factored, say it turns out to be something like (2x-1)(x+3)(x-2). Set this equal to 0, and solve for x
$\displaystyle (2x-1)(x+3)(x-2) = 0$ This is only true when any one of the factors is 0,so you have
$\displaystyle 2x-1=0$
$\displaystyle x+3=0$
$\displaystyle x-2=0$

Solving these for x, we see that $\displaystyle x = 1/2, -3, 2$

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# polynomial px qx has remainder x 1 when divided x^2-3 2

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