# Thread: Polynomials: Synthetic Division. Help needed ASAP!!

1. ## Polynomials: Synthetic Division. Help needed ASAP!!

Hello, my cousin has a math problem and needs this problem solved soon. I myself am not sure so any help would be great!

f(x)= x cubed - x squared -5x -3

a) i) Show that (x+1) is a factor of f(x)
ii) Hence or otherwise factorise f(x) fully

b) One of the turning points of the graph of y=f(x) lies on the x-axis. Write down the coordinates of this turning point

It is part b which she needs help with

I typed part a out incase it is needed.

2. You know that it has x+1 as a factor. Completed, the factorization is

$(x-3)(x+1)^{2}$

The turning point that lies on the x-axis is where it has a root of multiplicity 2.

See which one that is?.

3. I'm sorry but I am not sure what you mean

purely an educated guess but is it (1,-3)?

Possibly? Really not sure thanks for the reply

4. $f(x) = x^3 - x^2 - 5x - 3$

A)
i. To show that x+1 is a factor of f(x), you can just use long division and show that it divides without leaving any remainder.
ii. So you can factorize it as (x+1)(quotient). If it needs, you can factorize the quotient too.

B) It says that one of the turning points lies on x axis. First, find the turning points using $f'(x) = 0$. You should find two turning points. One of them is on the x axis, plug them in f(x) and see which one is.

5. Ahh!
Theres hope for me and my cousin yet!!

so you get, x= 5/3 and x=-1 , plug it so you get f(x)=0 ??

Thanks!

6. Hello, Scotland!

$f(x)\:=\: x^3 -x^2-5x - 3$

a) i) Show that $(x+1)$ is a factor of $f(x)$
. .ii) Factorise $f(x)$ fully.
We can use the Remainder/Factor Theorem.

. . $f(\text{-}1) \:=\\text{-}1)^3 - (\text{-}1)^2 - 5(\text{-}1) - 3 \;=\;0" alt="f(\text{-}1) \:=\\text{-}1)^3 - (\text{-}1)^2 - 5(\text{-}1) - 3 \;=\;0" />

Since $f(\text{-}1) = 0$, then $(x+1)$ is a factor of $f(x).$

Then we have: . $f(x) \;=\;(x+1)(x^2-2x-3) \;=\;(x+1)^2(x-3)$

b) One of the turning points of the graph of $f(x)$
lies on the x-axis. .Find the coordinates of this turning point.
We can answer this without Calculus . . .

In part (a), we found that: . $f(x) \:=\x+1)^2(x-3)" alt="f(x) \:=\x+1)^2(x-3)" />

The curve has x-intercepts at $(-1,\,0)\text{ and }(3,\,0)$.

The intercept $(-1,\,0)$ has multiplicity 2.
. . Hence, the curve is tangent to the x-axis at $x = -1.$

Therefore, there is a turning point at $(-1,\,0).$

The graph looks like this:
Code:
             |
|          *
|
|
|         *
|
--------o----+--------o----
* -1  * |       * 3
*         *     *
*          |  *
|
*           |

7. You don't necessarily need calculus. See where it touches but does not cross the x-axis?. Which root is that?. It's in your factorization. Which one has exponent 2?.

8. Ah right

A question, why is (-1,0) a tangent because it has multiplicity of 2?

Don't really understand that part...

Thanks again

9. $f(x) = (x+1)^2(x-3)$

Think about the graph y = x+1. The root x = -1 has multiplicity one because it can be written $y = (x+1)^1$. It crosses the x-axis there because it is negative on one side and positive on the other.

If we have a root of multiplicity 2, we get something like $y = (x+1)^2$. But $(x+1)^2$ is never negative so instead of crossing the x-axis at x = -1, it just touches it, then moves off in the direction it came from. So when we have roots of multiplicity 2, they have a tangent at the axis.

This can be generalised to all even and odd multiplicities.

10. Oh ok, I understand now.