Question

One factor of \[P(x)=4x^3+8x^2-ax-b\] is x+2. When the polynomial is divided by x-1, the remainder is -15. Find a and b, and hence factor the polynomial completely.

Answer 1

I think I would have been okay if it was one term... but two terms is confusing me on what I should do first...

Answer 2

@Limitless Do you know how to answer this question?

Answer 3

Ok.. So, do you know synthetic division?

Answer 4

Yes... but I don't know where to use it...

Answer 5

I think this will take me a moment. Just a sec.

Answer 6

Okay :).

Answer 7

@limitless did you get stuck? @satellite73 Do you know how to do this?

Answer 8

I am quite confused by this problem. I may not arrive at a solution.

Answer 9

Sigh... and I am being tested on this in the morning... :'(

Answer 10

Disregard that, I (am so tempted to use a meme here) was overthinking.

\[\frac{P(x)}{x+2}=4x^2-\frac{ax+b}{x+2}\]
Since \(x+2\) is a factor, it implies \(a=1\) and \(b=2\).

Does this make sense?

Answer 11

I apologize! It should be expressed
\[\frac{P(x)}{x+2}=4x^2+\frac{-ax-b}{x+2},\]
which implies \(a=-1\) and \(b=-2.\)

Answer 12

Not at all... lol

Answer 13

:\...why did you express it that way?

Answer 14

\(p(-2)=0\) and \(p(1)=-15\) you get two equations in two unknowns

Answer 15

Okay, here is why that logic works. When we say something is a factor of a polynomial, it must divide in a way such that there is no rational function. So we can't have the expression
\[\frac{-ax-b}{x+2}\]
in our division because this is a rational function. So, we must find the \(a\) and \(b\) such that \(-ax-b=x+2\). This is how I got that \(a=-1\) and (b=-2\).

Answer 16

although i assume @Limitless is correct, i would solve two equations

Answer 17

@satellite73 how does that allow me to solve for a and b? and @Limitless I get confused when math is in two much words...

Answer 18

replace \(x\) by 1, set result equal to \(-15\) you get
\[4+8-a-b=-15\]
\[12-a-b=-15\]
\[-a-b=-27\] or \[a+b=27\]as one equation (check my arithmetic)

Answer 19

The statement \[\frac{P(x)}{x+2}=4x^2+\frac{-ax-b}{x+2}\]
comes from polynomial division. The fraction part is called the remainder. You cannot have a remainder in polynomial division if what you are dividing by is a factor of your polynomial.

Because \(x+2\) is a factor, \(x+2\) should divide \(-ax-b\). Therefore, we say.
\[-ax-b=x+2.\]

Now, the only way that \(-ax-b\) is equal to \(x+2\) is if \(a=-1\) and \(b=-2.\) You can see this as follows
\[-ax-b=-(-1)x-(-2)=x+2.\]

Answer 20

replace \(x\) by 2 get \[4\times 2^3+8\times 2^2-2a-b=0\]
\[64-2a-b=0\]
\[-2a-b=-64\]
\[2a+b=64\]

Answer 21

solve the equations
\[2a+b=64\]
\[a+b=27\] and if my arithmetic is correct this should work quickly

Answer 22

@satellite73, that is giving a different answer.. Hmm.

Answer 23

how?

Answer 24

@Limitless did i mess up or is this arithmetic correct?

Answer 25

I believe my answer is correct because the division works out in both cases. You must have messed up the arithmetic.

Answer 26

it is late and hard for me to see the question and answer at the same time
but i think this is the easiest method. you have two equations in two unknowns, should work quickly

Answer 27

yeah i screwed up and used 2 instead of \(-2\) lets to it right

Answer 28

@satellite73, I was wrong, actually. I messed up signs.

Answer 29

@Limitless and @satellite73 so how do I find a and b your way?

Answer 30

\[p(1)=-15=8+4-a-b\implies a+b=27\]
\[p(-2)=4(-2)^3+8(-2)^2-2a-b=0\]
\[-32+32-2a-b=0\]
\[-2a-b=0\]
\[2a+b=0\] we get two equations
\[a+b=27\]
\[2a+b=0\]

Answer 31

subtract first equation from second get \(a=-27\) and so \(b=54\)

Answer 32

That works, Sat. My method was incorrect.

Answer 33

i didn't check to see if it works, but recall that \(a=-27\) so \(-a=27\) and further \(b=54\) so \(-b=-54\) and your polynomial should be
\[p(x)=4x^3+8x^2+27x-54\] now we can check it

Answer 34

Actually.. no, it doesn't.
\(x+2\) does not divide \(P(x)\).
http://www.wolframalpha.com/input/?i=%284x%5E3%2B8x%5E2%2B27x-54%29%2F%28x%2B2%29

Answer 35

@purplec16, are you 100% sure there was not a typo? My method works if the remainder is \(15\), not \(-15.\)

Answer 36

no there is not a typo... to my knowledge...

Answer 37

But I'll ask...

Answer 38

ok lets start from the top, but this time maybe we can write it as \[p(x)=4x^3+8x^2+ax+b\] to get rid of the annoying minus signs. obviously we just want \(a\) and \(b\)

Answer 39

now you are told that \(x+2\) divides \(p(x)\)which means \(p(-2)=0\) so we replace \(x\) by -2, set the result equal zero and solve
\[p(-2)=4(-2)^3+8(-2)^2+2a+b=0\]
\[2a+b=0\]

Answer 40

Okay.

Answer 41

and you are told than when you divide by \(x-1\) you get \(-15\) which makes \(p(1)=-15\) we get
\[4+8+a+b=-15\]
\[12+a+b=-15\]
\[a+b=-27\]

Answer 42

two equations are
\[2a+b=0\]
\[a+b=-27\] subtract and get
\[a=27, b=-54\] did i mess up yet?

Answer 43

Well...before it was a=-27 and b=54

Answer 44

oh i am an IDIOT!

Answer 45

No you're not...

Answer 46

\[\frac{P(x)}{x+2}=G(x)=4x^2+a+\frac{b-2a}{x+2}\]
\[\frac{P(x)}{x-1}=H(x)=4x^2+12+a\frac{b-a+12}{x-1}\]
\[b-2a=0\]
\[b-a+12=-15\]
\[b-a=-27\]
...

Answer 47

\[-2a+b=0\]!!!

Answer 48

ok i forgot that \(x=-2\) as the coefficient on the \(a\) term.
god help me

Answer 49

Aha! Yes...

Answer 50

\[b-2-(b-a)=27 \Rightarrow -2+a=27 \Rightarrow a=29\]
\[b-29=-27 \Rightarrow b=3\]
Correct?

Answer 51

It is really time for bed, smh

Answer 52

\[b=2\]!!!

Answer 53

Satellite, I cannot add. Do not worry.

Answer 54

I really... don't understand how you came up with that @Limitless I am really new to this formula...

Answer 55

And I forgot how to solve simultaneous equations..., smh

Answer 56

@purplec16, we are both failing. This is us trying to work it out before explaining it.

Answer 57

\(-2a+b=0\)
\[a+b=-27\] subtract and get \(-3a=27\) so \(a=-9\) and \(b=18\) lets try that one

Answer 58

Yeah, it should've been \(a+b+12=-15\).

Answer 59

b=-18 you mean?

Answer 60

yeash

Answer 61

where'd the 12 come from? @Limitless

Answer 62

Up above, polynomial division.

Answer 63

http://www.wolframalpha.com/input/?i=%284x^3%2B8x^2-9x-18%29%2F%28x%2B2%29

Answer 64

And how do I factor the polynomial completely...

Answer 65

We have finally accomplished this ;_;

Answer 66

well you know it factors as \((x+2)(\text{something})\) i know right? arithmetic hell

Answer 67

Yes

Answer 68

you can find the "something" by
polynomial division
synthetic division
thinking

your choice although at this point thinking doesn't seem to be the best idea

Answer 69

i would recommend synthetic division

Answer 70

Thinking has demonstrated its own failings today.

Answer 71

And what does the wolframalpha link show me?

Answer 72

Okay I'll use synthetic

Answer 73

but you know \[4x^2+8x^2-9x-18=(x+2)(ax^2+bx+c)\] and clearly \(a=4\) and \(c=-9\)

Answer 74

so you only need to find \(b\) which you can do by thinking
or rather should be able to

Answer 75

yeah thinking did not serve me that well either

Answer 76

Isn't it 4x^3+8x^2+9x+18?

Answer 77

no i rewrote at \[4x^3+8x^2+ac+b\] because the minus signs were unnecessary and getting on my nerves

Answer 78

it is
\[4x^3+8x^2-9x-18=(x+2)(4x^2-9)=(x+2)(2x-3)(2x+3)\] it turns out \(b=0\) i this case

Answer 79

Lol... okay... Thanks so much both @Limitless and @satellite73 idk who to give the medal to... :\ but you were both of such great help... I hope this is right, it sure looks so :)). Goodnight :).

Answer 80

And I got that @satellite73

Answer 81

lol we gave one to each other! so no need

Answer 82

It is right, don't worry. You are welcome. And yeah, lol. We're sharing on this question!

Answer 83

Okay :). Lol, thanks because I would have felt really bad if I only gave one and you both were equally helpful. Aww, that's so sweet. Thanks again!!!