Given f(x) = x^3 + 4x -5, show x-1 is a factor

[DeanBH]

The polynomial f(x) is given by f(x) = x^3 + 4x -5

show x-1 is factor :

f(1)=0
1+4-5=0 Thus it's a factor.

express in form (x-1)(x^2+px+q)

ends up being X^2 + X + 5.

next question is :

hence show the equation f(x) = 0 has one real root state it's value.

i know i have to use b^2 -4ac. If i do that it's -19 which is <0 so it's 1 value. But how do I find the value. And why does all this stuff work.

My teacher is one that just tells you how to do it. I have no idea why I'm doing it and it makes no sense so it's harder to remember. I would get 1or 2 marks out of 3 for this.

Firstly i don't know how to find out what the real number is. and i don't know why this works.

explain plez?! 8D

 

[HallsofIvy]

? Didn't you start off by showing that f(1)= 0??

And you are trying x so that f(x)= 0??

 

[CrazyIvan]

First a short lesson in discriminants, since you asked why you use them.

Remember the equation for the roots of any polynomial $$f \left(x \right) = a x^{2} + bx + c$$ is
$$x = \frac{-b + \sqrt{b^{2} - 4 a c}}{2a}$$ and $$x = \frac{-b - \sqrt{b^{2} - 4 a c}}{2a}$$

The term $$b^{2} - 4 a c$$ is called the "discriminant". If the discriminant is negative, then in you would be taking the square root of a negative number, which you cannot do with the real numbers.


Now, how many roots must $$f \left(x \right) = x^{3} + 4 x - 5$$ have?
Do you know the value of any of the roots already?

 

[tiny-tim]

Firstly i don't know how to find out what the real number is …

Hi DeanBH!

I think you're getting confused … the real number is 1, from the (x - 1) factor that you've already found.

When the question says "show the equation f(x) = 0 has one real root", it means exactly one real root (and two complex roots), instead of three real roots. That's all!

 

[tron_2.0]

by working backwards, you can just show that x^3+4x-5 factors into (x^2+5)(x-1)

not sure if your teacher wants you to use the quadratic formula or not, but working backwards works just fine in some cases.

 

[HallsofIvy]

No, it doesn't.

(x²+ 5(x-1)= x³- x²+ 5x- 5, NOT x³+ 4x- 5. Did you mean (x²+ x+ 5)(x- 1)?

 

[wantanabe_bk]

i don't think it's is a difficult question ? .

 

[HallsofIvy]

I believe that's already been pointed out several times!

 

[tron_2.0]

No, it doesn't.

(x²+ 5(x-1)= x³- x²+ 5x- 5, NOT x³+ 4x- 5. Did you mean (x²+ x+ 5)(x- 1)?

gah my bad, yes i ment (x²+ x+ 5)(x- 1)