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difficulty..........please explain
- Thread starter sujoy
- Start date
G'day,
Are you familiar with the remainder/factor theorums?
If \(\displaystyle P(x) = x^3 - x^2 - 4\)
Then if \(\displaystyle P(2) = 0\) then (x-2) is a factor ?
No doubt Denis will make comment: treat it as a difference of two cubes perhaps?
Are you familiar with the remainder/factor theorums?
If \(\displaystyle P(x) = x^3 - x^2 - 4\)
Then if \(\displaystyle P(2) = 0\) then (x-2) is a factor ?
No doubt Denis will make comment: treat it as a difference of two cubes perhaps?
No comments: your dexteritial leger-de-main left yours truly speechless :roll:Unco said:G'day,
Are you familiar with the remainder/factor theorums?
If \(\displaystyle P(x) = x^3 - x^2 - 4\)
Then if \(\displaystyle P(2) = 0\) then (x-2) is a factor ?
No doubt Denis will make comment: treat it as a difference of two cubes perhaps?
Act 2, scene 3 (enter sujoy, Denis)
sujoy:
"x^3 - x^2 - 4 =0
(x-2)(x^2+x+2)=0
even after expansion I am not getting it "
Denis:
What meanest thou, oh sujoy?
sujoy:
oh woebegones, said expansion I cannot computeth and terminateth with original
Denis:
multiply the damn thing again! (x-2) times (x^2+x+2) DOES EQUAL x^3-x^2-4 :!:
sujoy:
et tu, Brute, and play it again, Sam
(Exit sujoy, Denis)
Hello, sujoy!
Unco's suggestion is correct,
. . but I found an off-the-wall way to factor it.
. . We have: \(\displaystyle x^3\,-\,2x^2\,+\,x^2\,-\,4\:= \:0\)
. . Factor: \(\displaystyle x^2(x\,-\,2)\,+\,(x\,-\,2)(x\,+\,2)\:= \:0\)
. . Factor: \(\displaystyle (x\,-\,2)(x^2\,+\,x\,+\,2)\;= \:0\)
[Kids, don't try this at home.]
Unco's suggestion is correct,
. . but I found an off-the-wall way to factor it.
Subtract and add \(\displaystyle x^2:\;\;x^3\,-\,x^2\,-\,4\,{\bf-\,x^2\,+\,x^2} \:= \:0\)step [1]: \(\displaystyle x^3\,-\,x^2\,-\,4\:=\:0\)
step [?]
step [2]: \(\displaystyle (x\,-\,2)(x^2\,+\,x\,+\,2)\:=\:0]\)
. . We have: \(\displaystyle x^3\,-\,2x^2\,+\,x^2\,-\,4\:= \:0\)
. . Factor: \(\displaystyle x^2(x\,-\,2)\,+\,(x\,-\,2)(x\,+\,2)\:= \:0\)
. . Factor: \(\displaystyle (x\,-\,2)(x^2\,+\,x\,+\,2)\;= \:0\)
[Kids, don't try this at home.]
I was hoping Sujoy meant that after expansion he/she still could not see how to factorise it.Denis said:No comments: your dexteritial leger-de-main left yours truly speechless :roll:Unco said:G'day,
Are you familiar with the remainder/factor theorums?
If \(\displaystyle P(x) = x^3 - x^2 - 4\)
Then if \(\displaystyle P(2) = 0\) then (x-2) is a factor ?
No doubt Denis will make comment: treat it as a difference of two cubes perhaps?
Act 2, scene 3 (enter sujoy, Denis)
sujoy:
"x^3 - x^2 - 4 =0
(x-2)(x^2+x+2)=0
even after expansion I am not getting it "
Denis:
What meanest thou, oh sujoy?
sujoy:
oh woebegones, said expansion I cannot computeth and terminateth with original
Denis:
multiply the damn thing again! (x-2) times (x^2+x+2) DOES EQUAL x^3-x^2-4 :!:
sujoy:
et tu, Brute, and play it again, Sam
(Exit sujoy, Denis)