Question

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

a: x^3-3x^2-22x+24
b: x^3 - x^2 - 26x - 24
c: x^3 + x^2 - 26x + 24
d: x^3 +3x^2 + 14x - 24

Answer 1

you could use the zero product property in reverse, then expand

but since you're given the answer choices you can just check each one

Answer 2

all you have to do is plug each root into each equation

if you get 0 for each root on a particular equation, then you have found your answer

Answer 3

plug each root in for x?

Answer 4

for instance, in choice A, the equation is y = x^3-3x^2-22x+24

if -4 was a root, then plugging x = -4 into the equation *should* give you y = 0

if it doesn't give you y = 0, then x = -4 isn't a root

Answer 5

As @whpalmer4 said, your expression is equal to \((x-r_1)(x-r_2)(x-r_3)...(x-r_k)\) so you can plug in your roots and expand out to get your equation.

Answer 6

you have to do this with all 3 roots (for each answer choice)

Answer 7

im confused between reading his then yours

Answer 8

go with whichever way is easiest for you

Answer 9

Two different approaches. Both are valid. One of is practical only because you have a list of answer choices that you can try.

Answer 10

so is B correct?

Answer 11

what do you get when you plug in x = 1 into choice B?

Answer 12

no zero haha. -50 :o

Answer 13

so x = 1 is NOT a root of x^3 - x^2 - 26x - 24

which means you can eliminate B

Answer 14

see how that works?

Answer 15

the result has to be zero if that x value is a root

Answer 16

well had C to start out with and you said its not A

Answer 17

when did I say it's not A? I never said either way

Answer 18

well its not A because -4 doesnt work

Answer 19

Plug in roots into the equations; only one will consistently yield 0 for each root.

Answer 20

what do you get when you plug x = -4 into choice A

Answer 21

-50

Answer 22

im obviously doing it wrong because -4 doesnt make any of them zero for me

Answer 23

how are you typing it into your calculator

Answer 24

im writing it out on paper

Answer 25

is it possible to show us what you're writing down so we can see where you're going wrong

Answer 26

ok your not helping me. I knew how to do the equation I came on here because like I just said im doing something rong. Im plugging -4 in for the X there isnt that much more to explain

Answer 27

ok sry

Answer 28

just keep in mind that x^3 turns into (-4)^3 = -64

and something like x^2 turns into (-4)^2 = 16

Answer 29

please try not to be so rude next time

Answer 30

ive narrowed it down to c and D. and Im sorry If I came off As rude but you werent helping me,

Answer 31

is it C @eSpeX

Answer 32

I did not get C,.

Answer 33

I did, \((x-(-4))(x-1)(x-6)\)

Answer 34

then it must be D because A or B didnt work for me

Answer 35

When you foil this, what do you get?

Answer 36

\((x^2+3x-18x)(x-6)\)

Answer 37

oops, I meant, (x^2+3x-4)(x-6)

Answer 38

Let's try evaluating A again. You say it doesn't work for -4 for you?
\[(-4)^3-3(-4)^2-22(-4)+24=-64-3(16)-22(-4)+24=\]

Answer 39

Okay, work through with @whpalmer4 and if you still haven't an answer we can try again. :)

Answer 40

se Thats where I messed up I just got 0 for A for -4

Answer 41

You should get 0 if it is a root...

Answer 42

Try it with the other two roots as well. If you get 0 for them as we'll, then A would be your answer. The value of the polynomial is 0 at all roots.

Answer 43

for 1 in A I got 12? is that incorrect?

Answer 44

@whpalmer4

Answer 45

Afraid so...

Answer 46

\[(1)^3-3(1)^2-22(1)+24=1-3-22+24\]

Answer 47

So its A?

Answer 48

Only if you try all 3 and get 0 for each...

Answer 49

Do you agree that 1 gives you 0 in A?

Answer 50

Yes I solved it

Answer 51

Okay, how about 6?

Answer 52

\[(6)^3-3(6)^2-22(6)+24=216-108-132+24=\]

Answer 53

Yup Its A!

Answer 54

:-)

Answer 55

Thank you so much!

Answer 56

The other approach to this was to do polynomial multiplication, namely \[(x+4)(x-1)(x-6)=(x^2-x+4x-4)(x-6)\]\[=.x^3-6x^2-x^2+6x+4x^2-24x-4x+24\]\[=x^3-3x^2-22x+24 \]

Answer 57

Both offer significant opportunities for error if your algebra is mistake-prone :-)