Question

What polynomial has roots of -6, 1, and 4 ?

x3 - 9x2 - 22x + 24

x3 - x2 - 26x - 24

x3 + x2 - 26x + 24

x3 + 9x2 + 14x - 24

Answer 1

@whpalmer4

Answer 2

Here's a handy tip: the value of the polynomial will be 0 at a root.

Answer 3

Plug in x=-6 and evaluate. If you get something other than 0, move on to the next polynomial.

If you have multiple polynomials that give you 0 for x = -6, try x=1, and then x = 4.

Actually, probably easiest to do x = 1 first, because the arithmetic is easier!

Answer 4

simplify (x+6)(x-1)(x-4)=0

Answer 5

it's an okay assumption here, because the leading terms of all the choices have 1 as the coefficient, but it isn't as useful as a general tactic because it won't work if there was a constant factor as well.

All of these equations have those roots, for example :-)

\[2 x^3+2 x^2-52 x+48,\\3 x^3+3 x^2-78 x+72,\\4 x^3+4 x^2-104 x+96,\\5 x^3+5 x^2-130 x+120,\\6 x^3+6 x^2-156 x+144,\\7 x^3+7 x^2-182 x+168,\\8 x^3+8 x^2-208 x+192,\\9 x^3+9 x^2-234 x+216,\\10 x^3+10 x^2-260 x+240\]

Answer 6

@prettigirlrock0804 how are we doing?

Do you know about synthetic division and substitution yet?

Answer 7

@whpalmer4 yes i know what a synthetic division is

Answer 8

and sorry for not responding the other day my computer wasnt acting right

Answer 9

How about now? Do you have an answer now? :-)

Answer 10

hold on a sec let me work it out

Answer 11

do you have to do the synthetic division for all of them

Answer 12

You don't need to do any synthetic division here. You're trying to find the polynomial whose roots are -6, 1, and 4. Here are some approaches you can use:

1) build the polynomial in factored form which has roots -6, 1, 4, expand it (multiply it out), and compare with the answer choices.

2) evaluate each polynomial at x = -6, x = 1, x = 4 and make sure that all 3 produce a value of 0. If not, that polynomial does not have that set of roots.

Now, for 2), using synthetic substitution (which is just like synthetic division) makes the process of evaluating the polynomials at each of those 3 values much faster than doing something like \[(-6)^3-9(-6)^2-22(-6)+24 = \]

Do you need further explanation of either 1 or 2?

Answer 13

no

Answer 14

for the prob that you did when telling me about part 2 i got 264

Answer 15

Right. So that means that that polynomial does not have a root at x = -6. You don't need to do any further testing of roots with that polynomial, as it is not a possible correct answer.

Answer 16

ok

Answer 17

Actually, "right" only applies to the fact that you didn't get 0 for the result. I get -384.

\[(-6)^3 -9(-6)^2-22(-6)+24 = -216-324+132+24 = -384\]

Answer 18

so are you not suppose to get a zero

Answer 19

Here's how you do it with synthetic substitution:

-6 | 1 -9 -22 +24

-----------------------
1

-6 | 1 -9 -22 +24
-6
-----------------------
1 -15

-6 | 1 -9 -22 +24
-6 90
-----------------------
1 -15 68


-6 | 1 -9 -22 +24
-6 90 -408
-----------------------
1 -15 68 -384

Answer 20

You ARE supposed to get a 0 for f(x) if x is a root of f(x)

Answer 21

ok i got the final answer and its D

Answer 22

For example, let's take the polynomial \[(x+1)(x-2) = x^2 -2x + 1x -2= x^2-x-2\]

This has roots x = -1 and x = 2

If you plug in x = -1:
\[(-1)^2-(-1)-2) = 1+1-2 = 0\]so x=-1 is a root
If you plug in x = 1:
\[(1)^2-(1)-2 = 1-1-2 = -2\]so x = 1 is not a root

Answer 23

Really? What happens when you plug in x = 4?

\[ x^3 + 9x^2 + 14x - 24\]
\[(4)^3+9(4)^2+14(4)-24\]
That's going to equal 0?!?

Answer 24

yeah i just realized that

Answer 25

ugh this is making me mad

Answer 26

i tried all of them but yet cant find the correct answer

Answer 27

@whpalmer4 hello are you still there

Answer 28

Show me what you got when you evaluated the second polynomial

Answer 29

\[x^3+x^2-26x+24\]

Answer 30

-69

Answer 31

-6 1 1 -26 24
-6 30 -24
---------------
1 -5 4 0

so -6 is a root

1 1 1 -26 24
1 2 -24
--------------
1 2 -24 0

so 1 is a root

4 1 1 -26 24
4 20 -24
-------------
1 5 -6 0

so 4 is a root

Or doing it the hard way:

\[(-6)^3 + (-6)^2-26(-6)+24 = -216 + 36 +156 + 24 = 0\]
\[(1)^3+(1)^2-26(1)+24 = 1+1-26+24 = 0\]
\[(4)^3+(4)^2-26(4)+24 = 64+16-104+24=0\]

Again, all 3 numbers are roots.

Answer 32

I'm not sure which value of x you were using when you got -69, but you shouldn't get -69 for any of the 3 in that polynomial...if you show me your work, we can try to find the error so you don't make it or similar errors in the future.

Answer 33

ok

Answer 34

\[x ^{3}-x ^{2}-26x+24\]
\[-6^{3}-(-6)^{2}-26(-6)+24\]

Answer 35

Okay, one problem is that we did different polynomials :-) It appears I told you to try the second one, but then I actually used the third one myself. In my defense, I did show the polynomial I was suggesting...

Please try again with \[x^3+x^2-26+24\]

Also, be very careful when writing powers of negative numbers. \[-6^3 = -(6^3)\]though you get the same answer either way. However, \[-6^2 \ne (-6)^2\]and in both cases, the one with parentheses is the correct one.

Answer 36

i think i had messed up on my multiplication somewhere

Answer 37

Get in the habit of always writing ( ) around a negative number when raising it to a power, because if the power is an even number, you'll probably get the wrong answer if you don't.

Answer 38

So that should have been\[(-6)^3-(-6)^2-26(-6) + 24\]which evaluates to\[-216-36+156+24 = -72\]

Answer 39

Sigh. I made another typo. Please try again with \[x^3+x^2-26x+24\]:-)

Answer 40

yeah i found out where i have messed up

Answer 41

and ok

Answer 42

Did you see how my synthetic substitution thing made evaluation of the polynomials easier?

Answer 43

with the -6