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