# Math Help - f(x)=2x^3+13x^2+13x-10 i asked a question with this earlier but dnt understnd parts

1. ## f(x)=2x^3+13x^2+13x-10 i asked a question with this earlier but dnt understnd parts

f(x)=2x^3+13x^2+13x-10

find the value of f(2),f(-2). state one of the factors of f(x). Factorise f(x) completely.

can someone fully explain this?

thanks

2. find the value of f(2),f(-2).

Compare this with f(x). This means subbing x=2 into the bracket.

"state one of the factors of f(x)"
If it is a factor of f(x), it will not have any remainder.
therefore, for eg. when u sub x=-2 into f(x),
f(x)=2x^3+13x^2+13x-10
f(-2)=2{(-2)^3} +13{(-2)^2}+13(-2)-10=Remainder=0
This means (x+2) is a factor of f(x)

3. For finding other factors of f(x), i used the synthetic method.(which is a shortcut)

f(x)=(x+2)(ax^2+bx+c) This is standard

...in your original response earlier today the bit i dont udnerstand is

f(x)=(x+2)(ax^2+bx+c)

"by synthetic mtd" and the little table that followed

then 2x^2+9x-5 = ans what part of the question does this answer?

5. Originally Posted by maybeline9216
For finding other factors of f(x), i used the synthetic method.(which is a shortcut)

f(x)=(x+2)(ax^2+bx+c) This is standard

thats what i was after!! THANKS! your a life saver. i had basically done it. thanks for the rest of the help too your a star :]

6. Umm..u see..

f(x)=(x+2)(ax^2+bx+c)
f(x)=(x+2)(2x^2+9x-5)

Do u see any resemblance??

Actually, when im doing the synthetic mtd, im actually finding the "ax^2+bx+c" which comprises of other factors of f(x) because
(2x^2+9x-5)=(2x-1)(x+5)

Therefore, f(x)=(x+2)(2x-1)(x+5)

7. ## Synthetic method

could you explain the synthetic method to me as i struggle to factorise, its where i was stuck in the first place XD !

8. Originally Posted by coyoteflare
could you explain the synthetic method to me as i struggle to factorise, its where i was stuck in the first place XD !
As was stated before, since f(-2)=0, that makes x+2 a factor of $2x^3+13x^2+13x-10$.

Synthetic division works like this:

Code:

-2 |2  13    13  -10
|   -4   -18   10
| ---------------
2   9    -5    0
Since the dividend was degree 3, the quotient will be degree 2.

$2x^2+9x-5$

Thus, the two factors so far are: $(x+2)(2x^2+9x-5)$

The trinomial has already been factored for you in Post #6

9. where does the -18 and -4 come from? the divisend degree 3? what? please explain ..

10. Take whatever is the result on the bottom and multiply it by the -2 in the corner. I'll try to explain this a bit in more detail.

You start with 2, and it goes straight down. Then multiply that 2 by the -2 in the corner and put it into the next column, so you get -4. 13 + (-4) = 9, so you write that on the bottom. Then multiply it by the -2 in the corner to get -18, and put that into the next column. 13 + (-18) = 5, and you put that on the bottom. Continue this, and you get your answer.

11. Originally Posted by coyoteflare
where does the -18 and -4 come from? the divisend degree 3? what? please explain ..
You really need to refer to the link I gave you. There are other links out there as well. Just google "synthetic division" and you'll find many examples and explanations.

The dividend is made up of the coefficients of the original cubic expression.

The divisor comes from (x - c) being a factor, so c is my divisor. Specifically, (x+2) is a factor, by using a divisor of -2, the remainder will be 0.

First, I bring down the leading coefficient (2).
Second, I multiply my divisor (-2) by (2) to get (-4)
Third, I add (13) and (-4) to get (9).
Fourth, I multiply (-2) times (9) to get (-18).
Fifth, I add (13) to (-18) to get (-5)
Sixth, I multiply (-2) times (-5) to get (-10)
Finally, I add (10) to (-10) to get a remainder of (0).

The explanation on the link is much better than this. Look there.