# Thread: Polynomial functions. Thanks :)

1. ## Polynomial functions. Thanks :)

1. The function f(x) = 2x^3 +ax^2 -bx +3 has a factor (x+3). When f(x) is divided by (x-2), the remainder is 15.
a. Calculate the values of a and b
b. Find the other two factors of f(x)

2. Let P(x)=x^5-3x^4+2x^3-2x^2+3x+1
Given that P(x) can be written in the form (x^2-1)Q(x)+ax +b where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by x^2-1

3. The expressions px^4-5x+q and x^4-2x^3-px^2-qx-8 have a common factor x-2. Find the values of p and q

4. Write down one quadratic factor of x^4+x^3-x^2-3x-6, and find a second quadratic factor.

ps theses are questions from my textbook so i know wot the answers r i just can't get to them. if u don't might could u explain what ur doing.thanks.

2. Originally Posted by chaneliman
1. The function f(x) = 2x^3 +ax^2 -bx +3 has a factor (x+3). When f(x) is divided by (x-2), the remainder is 15.
a. Calculate the values of a and b
since (x + 3) is a factor, it means that f(-3) = 0 by the factor theorem

since when f(x) is divided by (x - 2) the remainder is 15, it means f(2) = 15 by the remainder theorem

this gives you two simultaneous equations from which you can solve for a and b

b. Find the other two factors of f(x)
once you find a and b, divide f(x) by (x + 3) using long or synthetic division. the quotient will be a quadratic from which you can find the other two factors. or you can try your luck with the rational roots theorem, but you've probably never seen that

3. Originally Posted by chaneliman
1. The function f(x) = 2x^3 +ax^2 -bx +3 has a factor (x+3). When f(x) is divided by (x-2), the remainder is 15.
a. Calculate the values of a and b
b. Find the other two factors of f(x)

2. Let P(x)=x^5-3x^4+2x^3-2x^2+3x+1
Given that P(x) can be written in the form (x^2-1)Q(x)+ax +b where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by x^2-1

3. The expressions px^4-5x+q and x^4-2x^3-px^2-qx-8 have a common factor x-2. Find the values of p and q

4. Write down one quadratic factor of x^4+x^3-x^2-3x-6, and find a second quadratic factor.

ps theses are questions from my textbook so i know wot the answers r i just can't get to them. if u don't might could u explain what ur doing.thanks.
1. This one isn't hard. When x = -3 then f(x) = 0 and when x = 2 then f(x) = 15. Solve the equations simultaneously.

2. $x = \pm 1$. Set those values into f(x)

3. When you set x = 2 then the expressions will both be equal to 0. This is another problem where you should use simultaneous equations.

4. Originally Posted by chaneliman
4. Write down one quadratic factor of x^4+x^3-x^2-3x-6, and find a second quadratic factor.
The rational root theorem fails to find a rational root for the quartic, so we are essentially left with some sort of "guess and check." However if we are to write a quadratic factor then we know that the polynomial factors as
$x^4+x^3-x^2-3x-6 = (x^2 + ax + b)(x^2 + dx + d)$
where a, b, c, and d are integers.

So let's expand out the right hand side and compare coefficients with the quartic. This gives the system of equations:
$a + c = 1$

$ac + b + d = -1$

$ad + bc = -3$

$bd = -6$

I think the easiest way to solve this is to solve the top and bottom equations for c and d respectively, giving
$-a^2 + a +b - \frac{6}{b} = -1$

$-a \left ( b + \frac{6}{b} \right ) + b = -3$

Now, if we are going to have integer values for b, then from bd = -6 we know that b can only b $\pm 1, \pm 2, \pm 3, \pm 6$.

From this I get solutions
a = 1, b = 2
a = 0, b = -3

So
$x^4+x^3-x^2-3x-6 = (x^2 + x + 2)(x^2 - 3)$

-Dan