Question

If x=2 is a root of x^3 + 3x^2 - 4x -12 =0 use synthetic division to factor the polynomial completely and find all zeros of the equation

Answer 1

Do you know how to do synthetic division?

Answer 2

Not really

Answer 3

If \(a\) is a root (or zero) of some polynomial \(P(x)\), then \(P(a) = 0\) and \[\frac{P(x)}{x-a}\]will have a remainder of \(0\).

In general, you can write a polynomial with zeros \(z_1, z_2, ... ,z_n\) as \[P(x) = a(x-z_1)(x-z_2)...(x-z_n)\]where \(a\) is a constant (used for causing the polynomial to go through a given point, or can be 1 if that isn't necessary). This means if you find a zero by hook or by crook, you can divide by \(x-zero\) to simplify your polynomial to one which has the same remaining zeros. Eventually, you end up with something that you can factor or solve.

Answer 4

Okay, when you do synthetic division, you write down the coefficients of the polynomial in descending order of their exponent, with a 0 for any value of the exponent which is missing. For example, if you had \(x^2-16\) you would write 1 0 -16 because you're missing the \(x\) term. Okay so far?

Answer 5

brb...

Answer 6

So I would have 1 3 4 12

Answer 7

no, 1 3 -4 -12

Answer 8

How do I draw on here.

Answer 9

Click on the blue Draw button

Answer 10

Where is that located.

Answer 11

right below where you type!

Answer 12

If \(x=2\) is a root, we'll divide by \((x-2)\) to simplify the polynomial.

Answer 13

Ok can't find the button, I'm on an ipad

Answer 14

Would the zeros be -1,0,6?

Answer 15

Oh, yes, no draw button on the iPad, I see (had never even tried to use it, and find OS a miserable experience on the iPad!)

Answer 16

we'll just do it this way:

1 3 -4 -12
2
-----------------------
1

We write 2 because we are dividing by (x-2) — we change the sign when constructing the table so that we can add rather than subtract. We drop the 1 down below the line because it is the 1st coefficient.

Now, we multiply 2 * 1 (the one below the line) and write the result above the line in the next column to the right:

1 3 -4 -12
2 2
-----------------------
1

Now we add down that column and write the result below the line:
1 3 -4 -12
2 2
-----------------------
1 5

Now we repeat the process:

1 3 -4 -12
2 2 2*5
-----------------------
1 5 6

and again:

1 3 -4 -12
2 2 2*5 2*6
-----------------------
1 5 6 0

So the result is that \[\frac{x^3+3x^2 -4x - 12}{x-2} = 1x^2 + 5x + 6\]

Answer 17

Can you factor that new polynomial?

Answer 18

Yes I get x+2)(x+3)

Answer 19

Okay, so here's the tricky bit: what are the zeros of that equation?

Answer 20

You may want to reread what I wrote in my second post before answering...

Answer 21

Would the answers be 0'-2,-3?

Answer 22

well, you didn't fall into the trap, but one of those numbers is not correct. Care to guess which one?

Answer 23

0?'

Answer 24

Yes, 0 is not a zero of this polynomial, because \(x^3+3x^2-4x-12 \ne 0 \text{ when }x=0\)

Answer 25

We successfully divided by \((x-2)\) and got a remainder of \(0\), so \((x-2)\) must be a factor of \[x^3+3x^2-4x-12 =0\]and for that to be the case, \((x-2) = 0\) will yield a zero of...

Answer 26

2' thank you soooooo much for helping me!!!

Answer 27

You're welcome!

Note that you can also use synthetic division to evaluate a polynomial quickly. Say we had to find the value of your polynomial when \(x=1\) (where it will not be \(0\), as \(x=1\) is not a zero of the polynomial:

You could laboriously do \[(1)^3+3(1)^2-4(1)-12 = 1+3-4-12 = -12\](well, okay, that isn't so laboriously, but it would be painful if we had to do \(x=-5\), for example)

Here's how you do synthetic substitution, which is just like synthetic division, except you interpret the result differently:
1 3 -4 -12
1 1 4 0
-----------------
1 4 0 -12

so the polynomial equals -12 at x = 1

1 3 -4 -12
-5 -5 10 -30
--------------
1 -2 6 -42

polynomial equals -42 at x = -5

sure beats doing \((-5)^3+3(-5)^2-4(-5)-12\) doesn't it?

Answer 28

I think you just helped me with another one of my problems! Have a great day !

Answer 29

Great! Enjoy your afternoon! :-)