hi there, just need some clarification on one point, i'll post the whole question and show you what i've done so far.

a. The polynomial 2x^3-5x^2+46x+87 has one real factor (2x+3) and two complex factors.

By algebraic division and solving a quadratic equation find the compex factors and express the above polynomial in the form 2x^3-5x^2+46x+87 = (2x+3)(x-a)(x-b) where a and b are complex numbers.

b. Calculate the gradient of the curve y=2x^3-5x^2+46x+87 at the point where it crosses the x axis

c. Show by differentiation and solving a quadratic equation that there are no points on the above curve where the gradient is 0.

The answers i got for a were (2x+3)(x-(2+j5)(x-(2-j5)

The answer i got for b was 74.5

Its part c that i'm not sure of, how can i tell the gradient never equals 0?

Any help is apprecciated!

2. Originally Posted by ally79
hi there, just need some clarification on one point, i'll post the whole question and show you what i've done so far.

a. The polynomial 2x^3-5x^2+46x+87 has one real factor (2x+3) and two complex factors.

By algebraic division and solving a quadratic equation find the compex factors and express the above polynomial in the form 2x^3-5x^2+46x+87 = (2x+3)(x-a)(x-b) where a and b are complex numbers.

b. Calculate the gradient of the curve y=2x^3-5x^2+46x+87 at the point where it crosses the x axis

c. Show by differentiation and solving a quadratic equation that there are no points on the above curve where the gradient is 0.

The answers i got for a were (2x+3)(x-(2+j5)(x-(2-j5)

The answer i got for b was 74.5

Its part c that i'm not sure of, how can i tell the gradient never equals 0?

Any help is apprecciated!
For c)

$\frac{dy}{dx} = 6x^2 - 10x + 46= 0$

By checking its discriminant, we see that $\Delta = (-10)^2 - 4(6)(46)$ which is clearly < 0.

Since the derivative can never be solved = 0, we can tell that the gradient never equals 0 (since the derivative is the gradient at any point).

3. Sorry it may just be me being a bit thick but is there any chance you can break it down a bit as I dont understand where the rest of the quadratic has gone.

Thanks

4. Originally Posted by ally79
Sorry it may just be me being a bit thick but is there any chance you can break it down a bit as I dont understand where the rest of the quadratic has gone.

Thanks
The solution to any quadratic of the form

$ax^2 + bx + c = 0$

is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. (Why?)

Notice that there is a term under a square root here. This is called the discriminant, and is denoted by $\Delta$.

If $\Delta > 0$ then there are two solutions for x.

If $\Delta = 0$ then there is one solution for x.

If $\Delta < 0$ then there does not exist any solutions for x.

In your problem, $a = 6, b = -10, c = 46$.

What does $\Delta = b^2 - 4ac$ equal in this case? What does this tell you about the solutions?

5. Thanks very much for that, didnt know about the discriminant but now understand.