# factorsWatch

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#1
find the co ordinates where the graph of

y=2x^3 + 3x^2 - 4x +1

cuts the x axis??????
0
14 years ago
#2
find the co ordinates where the graph of

y=2x^3 + 3x^2 - 4x +1

--

Factor theorem
F(x) = 2x^3 + 3x^2 - 4x +1
When F(1) = 2 + 3 - 4 + 1 = 2 therfore (x - 1) is not a factor
F(-1) = -2 + 3 + 4 + 1 = -2

Factor theorem doesn't work...

What question is this btw, c2 or c3?

you can only use the factor theorem for factors of 1 as 1 is the integer at the end.
0
14 years ago
#3
(0,1) isn't a solution because that's not on the x-axis.
also if your factorisation were correct, the curve would be y = x^3 + 3x^2 + 3x + 1 by the binomial expansion.

try searching for factors by the factor theorum.
0
14 years ago
#4
drawing a graph doesnt give a nice root.. its like -2.4 something
0
14 years ago
#5
Can't be a c2 question asthe factor theorem doesnt even work.

1 & -1 are both unsuccessful

you sure the question is right?

Could always try numerical methods
0
14 years ago
#6
(Original post by sarah12345)
find the co ordinates where the graph of

y=2x^3 + 3x^2 - 4x +1

cuts the x axis??????
f(x)=2x^3+3x^2-4x+1
f(1/2)=2/8+3/4-2+1=0
factor theorem gives (2x-1) is a factor
2x^3+3x^2-4x+1=(2x-1)(x^2+2x-1)
f cuts x-axis at x=1/2
x=-2+-rt(4+4)/2=-1+-rt(2)
0
#7
(Original post by evariste)
f(x)=2x^3+3x^2-4x+1
f(1/2)=2/8+3/4-2+1=0
factor theorem gives (2x-1) is a factor
2x^3+3x^2-4x+1=(2x-1)(x^2+2x-1)
f cuts x-axis at x=1/2
x=-2+-rt(4+4)/2=-1+-rt(2)
how did you get f(1/2) do you just guess?
0
14 years ago
#8
(Original post by sarah12345)
how did you get f(1/2) do you just guess?
sort of. this is an A level question so its not going to be too hard. i would suggest if 1,-1,2,-2 fail then try 1/2,-1/2.
0
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