Polynomials - Polynomial Equations

[TheDanny]

Trying to ask my method of doing the question is correct or not? Try check is there any mistake please? I am beginner.
The Question
Find the real solutions to each of the following questions.
(a) X³+X²-10X+8=0My attempt at a solution

X³+X²-10X+8=0
f(1)=1³+1²-10(1)+8
f(1)=0
X=1
X-1=0 (one of the factor)

X³+X²-10X+8=(X-1)(X²+kX-8)
Compare the coefficient X²
X²=kX²-X²
1=k-1
2=k

X²+X-8=(X-2)(X+4)
The factors are (X-1)(X-2)(X+4)
**(From here start my working wonder correct or wrong)***
1st Method of answering
(X-1)(X-2)(X+4)=0
X-2=0, X-1=0, X+4=0
X=2, X=1, X=-4 #
2nd Method of answering
f(X)=(X-1)(X-2)(X+4)
f(X)=0
(X-1)(X-2)(X+4)=0
X-2=0, X-1=0, X+4=0
X=2, X=1, X=-4 #

The 1st Method of answering or 2nd is correct?

 

[malawi_glenn]

X²+x-8=(x-2)(x+4) ?

You should try to learn to perform division of polynomials.

 

[symbolipoint]

TheDanny, exactly what course are you studying?

 

[Feldoh]

Both methods are correct, it's just implied the f(x) = 0 = (x-1)(x-2)(x+4)

 

[HallsofIvy]

Trying to ask my method of doing the question is correct or not? Try check is there any mistake please? I am beginner.
The Question
Find the real solutions to each of the following questions.
(a) X³+X²-10X+8=0


My attempt at a solution

X³+X²-10X+8=0
f(1)=1³+1²-10(1)+8
f(1)=0
X=1
X-1=0 (one of the factor)

Yes, it is true that f(1)= 0 and so x= 1 is a root and x-1 is a factor of x³+ x²- 10x+ 8

X³+X²-10X+8=(X-1)(X²+kX-8)
Compare the coefficient X²
X²=kX²-X²
1=k-1
2=k

X²+X-8=(X-2)(X+4)

?? I was under the impression that -2x+ 4x= 2x, not X. This can't possibly be true. Did you consider the possibility that x²+x-8= 0 has NO rational roots? If that is true, then you might consider either completing the square or using the quadratic equation to solve x²+ x- 8= 0.

[/quote]The factors are (X-1)(X-2)(X+4)[/quote]
NO, they aren't. That's your error

**(From here start my working wonder correct or wrong)***
1st Method of answering
(X-1)(X-2)(X+4)=0
X-2=0, X-1=0, X+4=0
X=2, X=1, X=-4 #
2nd Method of answering
f(X)=(X-1)(X-2)(X+4)
f(X)=0
(X-1)(X-2)(X+4)=0
X-2=0, X-1=0, X+4=0
X=2, X=1, X=-4 #

The 1st Method of answering or 2nd is correct?

 

[learningphysics]

X²+x-8=(x-2)(x+4) ?

You should try to learn to perform division of polynomials.

He accidentally wrote X²+x-8 instead of X²+2x-8. Other than that everything is correct.

TheDanny, your solution is fine (except for leaving out the 2 in the polynomial). Both ways of answering seem fine to me.

 

[TheDanny]

He accidentally wrote X²+x-8 instead of X²+2x-8. Other than that everything is correct.

TheDanny, your solution is fine (except for leaving out the 2 in the polynomial). Both ways of answering seem fine to me.

Yeah i found that i accidentally wrote X²+x-8 without the 2 inside.
HallsofIvy, maybe you right over here that
X²+2X-8=(X-2)(X+4), or i should write

X²+2X-8=0
(X-2)(X+4)=0?

I study Mathematics T that is A-level or Pre-U i think. Higher School Certificate (HSC). The HSC was the precursor to the GCE A levels in the UK, and is still the name of the pre-university examination in some states in Australia.
Future Maths is hard or? I wonder my teacher tell me that not all can learn only some who is talented only. I will need all your help next time, thanks.

 

[HallsofIvy]

Don't you just hate when that happens?

Here I was thinking "It's so hot, I'll have a nice gin and tonic before I turn in. Next thing I know, it's 8:00 AM. I'd probably better not have a gin and tonic now!

Yes, if x²+ 2x- 8= 0 then (x-2)(x+4)= 0 and the solutions are x= 2 and x= -4. You can, op, of course, check that by just doing the arithmetic:
2^{2[/aup]+ 2(2)-8= 4+ 4- 8= 0 and (-4)2+ 2(-4)+ 8= 16- 8-8= 0.}

 

[TheDanny]

hmm

HallsofIvy,
this is mine error?
[/quote]The factors are (X-1)(X-2)(X+4)[/quote]
NO, they aren't. That's your error

Should it be The factors are (X-1),(X-2),(X+4)? put commas between each factor?

 

[Feldoh]

Should it be The factors are (X-1),(X-2),(X+4)? put commas between each factor?

I think so -- If you were to list the factors of 12 you wouldn't say 1*2*3*4*6*12, you'd say 1,2,3,4,6,12.