# help , remainder theoremWatch

#1
given that f(x)= x^3 + x^2 + kx -4 is divisible by (x+2), show that k = -4
b)factorise f(x) completely
c)solve f(x)=(x-2)(x+1)
d) solve f(x)=(x+1)

okay so I get b to be (x+2)(x+1)(x-2)

but for c and d I am not sure what to do. The answers in the book are 2 or -1 for b and -1 and + or 1 root 5 for d

how on earth do you do c and d? thanks
0
quote
5 years ago
#2
given that f(x)= x^3 + x^2 + kx -4 is divisible by (x+2), show that k = -4
b)factorise f(x) completely
c)solve f(x)=(x-2)(x+1)
d) solve f(x)=(x+1)

okay so I get b to be (x+2)(x+1)(x-2)

but for c and d I am not sure what to do. The answers in the book are 2 or -1 for b and -1 and + or 1 root 5 for d

how on earth do you do c and d? thanks
f(x) = f(x) Maybe? So that (x+2)(x+1)(x-2) = (x-2)(x+1)
generally the question should read solve f(x) = 0 or even another number . . . this would get you the right answer for c but not d.
0
quote
5 years ago
#3
Remember that

So for 'c', if you want to solve

From 'b', you already know what f(x) is, so you would write it as

Does this make more sense in how to solve it?
0
quote
5 years ago
#4
given that f(x)= x^3 + x^2 + kx -4 is divisible by (x+2), show that k = -4
b)factorise f(x) completely
c)solve f(x)=(x-2)(x+1)
d) solve f(x)=(x+1)

okay so I get b to be (x+2)(x+1)(x-2)

but for c and d I am not sure what to do. The answers in the book are 2 or -1 for b and -1 and + or 1 root 5 for d

how on earth do you do c and d? thanks
You get

now you should be able to write down the equations representing

Note that the easiest way to solve these is by taking out common factors at the front, i.e (x-2)(x+1) for (c) and (x+1) for (d).
0
quote
#5
(Original post by claret_n_blue)
Remember that

So for 'c', if you want to solve

From 'b', you already know what f(x) is, so you would write it as

Does this make more sense in how to solve it?
so just expand brackets and solve as normal?
0
quote
5 years ago
#6
so just expand brackets and solve as normal?
You needn't expand the brackets. Look at my post above.
0
quote
5 years ago
#7
so just expand brackets and solve as normal?
You could expand but I think it might take longer. Try rearranging and then factorising again by taking out common factors.
0
quote
#8
(Original post by claret_n_blue)
Remember that

So for 'c', if you want to solve

From 'b', you already know what f(x) is, so you would write it as

Does this make more sense in how to solve it?
I am having trouble solving it.. would you mind walking me through? I am up to the point of taking out the common factors. thanks so much
0
quote
5 years ago
#9
I am having trouble solving it.. would you mind walking me through? I am up to the point of taking out the common factors. thanks so much
Show me what you have written. We're not supposed to just tell you the answers on this site.
0
quote
#10
(Original post by claret_n_blue)
Show me what you have written. We're not supposed to just tell you the answers on this site.
(x+1)(x-2)[(x+2)-1]

not sure how to solve
0
quote
5 years ago
#11
(x+1)(x-2)[(x+2)-1]

not sure how to solve
You're telling us that you can't simplify (x+2)-1 ?
0
quote
#12
(Original post by Indeterminate)
You're telling us that you can't simplify (x+2)-1 ?
-x -2 ...
0
quote
5 years ago
#13
(x+1)(x-2)[(x+2)-1]

not sure how to solve
Yeah, that's correct so far.

Don't forget that [(x + 2) -1] is just (x + 2 - 1) which you can do.
0
quote
5 years ago
#14
-x -2 ...
What?!

x+2 - 1 =....
0
quote
X

new posts

Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• University of Lincoln
Mini Open Day at the Brayford Campus Undergraduate
Wed, 19 Dec '18
• University of East Anglia
Fri, 4 Jan '19
• Bournemouth University
Wed, 9 Jan '19

### Poll

Join the discussion

Yes (281)
27.6%
No (737)
72.4%