Question

hey can i get help with this please factor theorem

Answer 1

The remainder theorem states, that if you have a polynomial Q(x) divided by x-a then Q(a) will give you the remainder value.
So, you were given a polynomial that is divisible by two binomials.
Now, what happens when a polynomial is divisible by other? is there any remainder?

Answer 2

well i think there should be a reminder

Answer 3

That's incorrect, let's view it with numbers.
If I divide 36 by 6, is there any remainder?

Answer 4

Factor (x-2) ==> P(2) = 0
Factor (x+3) ==> P(-3) = 0
Not Factor (x+1), Remainder -18 ==> P(-1) = -18

Answer 5

wat did @tkhunny do?

Answer 6

What he did was applying the definition of divisibility with the first two binomials.
When a polynomial is divisible by other then the remainder is zero (or no remainder at all).
Then on the third one, he applied the remainder theorem.

Answer 7

This will create a 3x3 system of equations, where you'll have to solve for m, n and k respectively.

Answer 8

can u plss show me i am not getting it !

Answer 9

yup nope was going to try to help you but looks to hard. Good luck :)

Answer 10

So, following his deduction, we will obtain three different evaluations for the given polynomial:
\[P(x)=2x^4+mx^3-nx^2-7x+k \]
We were given that the polynomial is divisible by \((x-2), (x+3)\) so that must mean that their division must be zero, and the "factor theorem" is a consequence of the Remainder theorem, so, if we solve the divisible binomials:
\[x-2=0 \iff x=2\]
\[x+3=0 \iff x=-3\]
Then:
\[P(2)=0 \]
\[P(-3)=0\]
So therefore, evaluating the polynomials:
\[P(2)=2(2)^4+m(2)^3-n(x)^2-7(2)+k=0\]
\[P(-3)=2(-3)^4+m(-3)^3-n(-3)^2-7(-3)+k=0 \]

Answer 11

ok

Answer 12

so like do we solve the 2 equations or ??

Answer 13

You have to apply the remainder theorem in order to obtain a third equation.
I quote: "...And leaves a remainder of -18 when divided by (x+1)"
So, the steps are analogical to what I did above, something will change, the whole P(-1) will not be equal zero but rather the remainder that was given to us.

Answer 14

oh ok so i get the part in which you have to equate the equation to the reminder but which equation do we do it for

Answer 15

Good, you would have obtained three equations where the variables are m,n and k.

Answer 16

no no no wait which equation do i equate to the reminder ???

Answer 17

These being:
\[P(2)=0\]
\[P(-3)=0\]
\[P(-1)=-18\]

Answer 18

ya ok

Answer 19

Okay, I had written to you above what the evaluations of P(2) and P(-3) were, so you do the same for P(-1):
\[2(2)^4+m(2)^3-n(2)^2-7(2)+k=0\]
\[2(-3)^4+m(-3)^3-n(-3)^2-7(-3)+k=0\]
\[2(-1)^4+m(-1)^3-n(-1)^2-7(-1)+k=-18\]

Answer 20

After you simplify the operations of the equations above, you'll obtain a 3x3 system of equations with three variables m, n and k.
Which I must assume you already know how to solve.

Answer 21

yea i do

Answer 22

Well, I will assume you can take over from here.

Answer 23

yea thank you @Owlcoffee i owe you !!!

Answer 24

This is a very messy problem calculation. At the risk of being banned for life:
\[\left\{m\to \frac{18}{5},n\to \frac{39}{5},k\to -\frac{78}{5}\right\} \]

Answer 25

lol no thank you though