help , remainder theorem

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 :smile:

Reply 1

joostan

13

Original post by madfish

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 :smile:

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.

(edited 11 years ago)

Reply 2

claret_n_blue

17

Remember that

f(x) x^3 + x^2 - 4x - 4 = (x + 2) (x + 1)(x -2 )

So for 'c', if you want to solve

f(x) = (x -2 ) (x + 1)

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

(x + 2)(x + 1)(x - 2) = (x - 2)(x + 1)

Does this make more sense in how to solve it?

Reply 3

Indeterminate

3

Original post by madfish

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 :smile:

You get

f(x) = (x+2)(x+1)(x-2)

now you should be able to write down the equations representing

f(x)=(x-2)(x+1)

f(x)=(x+1)

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).

Reply 4

madfish

OP

Original post by claret_n_blue

Remember that

f(x) x^3 + x^2 - 4x - 4 = (x + 2) (x + 1)(x -2 )

So for 'c', if you want to solve

f(x) = (x -2 ) (x + 1)

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

(x + 2)(x + 1)(x - 2) = (x - 2)(x + 1)

Does this make more sense in how to solve it?

so just expand brackets and solve as normal?

Reply 5

Indeterminate

3

Original post by madfish

so just expand brackets and solve as normal?

You needn't expand the brackets. Look at my post above.

Reply 6

claret_n_blue

17

Original post by madfish

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.

Reply 7

madfish

OP

Original post by claret_n_blue

Remember that

f(x) x^3 + x^2 - 4x - 4 = (x + 2) (x + 1)(x -2 )

So for 'c', if you want to solve

f(x) = (x -2 ) (x + 1)

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

(x + 2)(x + 1)(x - 2) = (x - 2)(x + 1)

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 :colondollar:

Reply 8

claret_n_blue

17

Original post by madfish

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 :colondollar:

Show me what you have written. We're not supposed to just tell you the answers on this site.