anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
plug each root in for x?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
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
eSpeX
  • eSpeX
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.
jim_thompson5910
  • jim_thompson5910
you have to do this with all 3 roots (for each answer choice)
anonymous
  • anonymous
im confused between reading his then yours
jim_thompson5910
  • jim_thompson5910
go with whichever way is easiest for you
whpalmer4
  • whpalmer4
Two different approaches. Both are valid. One of is practical only because you have a list of answer choices that you can try.
anonymous
  • anonymous
so is B correct?
jim_thompson5910
  • jim_thompson5910
what do you get when you plug in x = 1 into choice B?
anonymous
  • anonymous
no zero haha. -50 :o
jim_thompson5910
  • jim_thompson5910
so x = 1 is NOT a root of x^3 - x^2 - 26x - 24 which means you can eliminate B
jim_thompson5910
  • jim_thompson5910
see how that works?
jim_thompson5910
  • jim_thompson5910
the result has to be zero if that x value is a root
anonymous
  • anonymous
well had C to start out with and you said its not A
jim_thompson5910
  • jim_thompson5910
when did I say it's not A? I never said either way
anonymous
  • anonymous
well its not A because -4 doesnt work
anonymous
  • anonymous
Plug in roots into the equations; only one will consistently yield 0 for each root.
jim_thompson5910
  • jim_thompson5910
what do you get when you plug x = -4 into choice A
anonymous
  • anonymous
-50
anonymous
  • anonymous
im obviously doing it wrong because -4 doesnt make any of them zero for me
jim_thompson5910
  • jim_thompson5910
how are you typing it into your calculator
anonymous
  • anonymous
im writing it out on paper
jim_thompson5910
  • jim_thompson5910
is it possible to show us what you're writing down so we can see where you're going wrong
anonymous
  • anonymous
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
jim_thompson5910
  • jim_thompson5910
ok sry
jim_thompson5910
  • jim_thompson5910
just keep in mind that x^3 turns into (-4)^3 = -64 and something like x^2 turns into (-4)^2 = 16
jim_thompson5910
  • jim_thompson5910
please try not to be so rude next time
anonymous
  • anonymous
ive narrowed it down to c and D. and Im sorry If I came off As rude but you werent helping me,
anonymous
  • anonymous
is it C @eSpeX
eSpeX
  • eSpeX
I did not get C,.
eSpeX
  • eSpeX
I did, \((x-(-4))(x-1)(x-6)\)
anonymous
  • anonymous
then it must be D because A or B didnt work for me
eSpeX
  • eSpeX
When you foil this, what do you get?
eSpeX
  • eSpeX
\((x^2+3x-18x)(x-6)\)
eSpeX
  • eSpeX
oops, I meant, (x^2+3x-4)(x-6)
whpalmer4
  • whpalmer4
Let's try evaluating A again. You say it doesn't work for -4 for you? \[(-4)^3-3(-4)^2-22(-4)+24=-64-3(16)-22(-4)+24=\]
eSpeX
  • eSpeX
Okay, work through with @whpalmer4 and if you still haven't an answer we can try again. :)
anonymous
  • anonymous
se Thats where I messed up I just got 0 for A for -4
whpalmer4
  • whpalmer4
You should get 0 if it is a root...
whpalmer4
  • whpalmer4
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.
anonymous
  • anonymous
for 1 in A I got 12? is that incorrect?
anonymous
  • anonymous
@whpalmer4
whpalmer4
  • whpalmer4
Afraid so...
whpalmer4
  • whpalmer4
\[(1)^3-3(1)^2-22(1)+24=1-3-22+24\]
anonymous
  • anonymous
So its A?
whpalmer4
  • whpalmer4
Only if you try all 3 and get 0 for each...
whpalmer4
  • whpalmer4
Do you agree that 1 gives you 0 in A?
anonymous
  • anonymous
Yes I solved it
whpalmer4
  • whpalmer4
Okay, how about 6?
whpalmer4
  • whpalmer4
\[(6)^3-3(6)^2-22(6)+24=216-108-132+24=\]
anonymous
  • anonymous
Yup Its A!
whpalmer4
  • whpalmer4
:-)
anonymous
  • anonymous
Thank you so much!
whpalmer4
  • whpalmer4
The other approach to this was to do polynomial multiplication, namely \[(x+4)(x-1)(x-6)=(x^2-x+4x-4)(x-6)\]\[=.x^3-6x^2-x^2+6x+4x^2-24x-4x+24\]\[=x^3-3x^2-22x+24 \]
whpalmer4
  • whpalmer4
Both offer significant opportunities for error if your algebra is mistake-prone :-)

Looking for something else?

Not the answer you are looking for? Search for more explanations.