help , remainder theorem Watch

madfish
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#1
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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
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joostan
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(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
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.
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claret_n_blue
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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?
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Indeterminate
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(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
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).
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madfish
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(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?
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Indeterminate
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(Original post by madfish)
so just expand brackets and solve as normal?
You needn't expand the brackets. Look at my post above.
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claret_n_blue
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(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.
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madfish
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(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
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claret_n_blue
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(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
Show me what you have written. We're not supposed to just tell you the answers on this site.
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madfish
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(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
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Indeterminate
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(Original post by madfish)
(x+1)(x-2)[(x+2)-1]

not sure how to solve
You're telling us that you can't simplify (x+2)-1 ?
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madfish
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#12
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(Original post by Indeterminate)
You're telling us that you can't simplify (x+2)-1 ?
-x -2 ...
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claret_n_blue
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(Original post by madfish)
(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.
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Indeterminate
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#14
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(Original post by madfish)
-x -2 ...
What?!

x+2 - 1 =....
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