Polynomials and Remainder Theorem (1 Viewer)

P!xel · Feb 15, 2009

This time, there's only 1 questions haha.

Question 1)- I remembered how,so ignore this one

.

Question 2)- Same goes with this one.

Question 3)

gurmies · Feb 15, 2009

Question 1:

$p(x) = x^{3} + ax^{2} + bx - 18$

$p(-2) = 0$

$\therefore (-2)^{3} + a(-2)^{2} + b(-2) - 18 = 0$

$4a - 2b = 26 \, \mathbf{(1)}$

$2a - b = 13\, \mathbf{(1)}$

$p(1) = -24$

$\therefore (1)^{3} + a(1)^{2} + b(1) - 18 = -24$

$a + b = -7 \, \mathbf{(2)}$

$Adding \, \mathbf{(1)} \, and \, \mathbf{(2)}$

$3a = 6$

$a = 2$

$\therefore b = -9$

Question 3:

(a)

$f(x) = x^{3} + 3x^{2} - 9x - 27$

$f(3) = (3)^{3} + 3(3)^{2} - 9(3) - 27 = 0$

$\therefore (x-3) \, is \, a \, factor \, of \, f(x) = x^{3} + 3x^{2} - 9x - 27$

$f(x) = x^{2}(x+3) - 9(x+3)$

$f(x) = (x+3)(x^{2}-9)$

$f(x) = (x-3)(x+3)^{2}$

(b)

$When \, f(x) = 0 \, , \, x = \pm 3$

(c)

$At \, stationary \, points \,\,\,\, \frac{dy}{dx} = 0$

$\frac{dy}{dx} = 3x^{2} + 6x - 9 = x^{2} + 2x - 3$

$\frac{d^{2}y}{dx^{2}} = 2x + 2$

$(x+3)(x-1) = 0$

$x = -3 \, or \, x = 1$

$\frac{d^{2}y}{dx^{2}} = 2x + 2$

$When \, x = -3, \,\,\,\,\,\,y = 0,\,\,\, \frac{d^{2}y}{dx^{2}} \, is \, negative$

$\therefore (-3, 0) \, is \, a \, maximum \, turning \, point$

$When \, x = 1, \,\,\,\,\,\,y = -32,\,\,\, \frac{d^{2}y}{dx^{2}} \, is \, positive$

$\therefore (1,-32) \, is \, a \, minimum \, turning \, point$

(d)

$At \, inflexion \, points, \, \frac{d^{2}y}{dx^{2}} = 0$

$2 + 2x = 0$

$x = -1$

$When \, x = -1, \, y = -16 \, and \, there \, is \, a \, change \, in \, sign \, through \, this \, point \, when \, plugged \, into \, the \, second \, derivative.$

$\therefore (-1, -16) \, is \, an \, oblique \, point \, of \, inflexion$

kurt.physics · Feb 15, 2009

Question 3)

[/quote]

a) if x-3 is a factor, then p(3) = 0. then just divide the polynomial.

b) make this factored expression from A equal 0 to find where it crosses the x axis and let x=0 to find where it crosses the y axis

c) calculus

d) calculus.

P!xel · Feb 15, 2009

kurt.physics said:
a) if x-3 is a factor, then p(3) = 0. then just divide the polynomial.

b) make this factored expression from A equal 0 to find where it crosses the x axis and let x=0 to find where it crosses the y axis

c) calculus

d) calculus.

I followed your instructions for a) and b) and it worked

.

But I forgot how to find the turning points..

(Did it have something to do with f''(x)?)

tacogym27101990 · Feb 15, 2009

P!xel said:
I followed your instructions for a) and b) and it worked .

But I forgot how to find the turning points..

(Did it have something to do with f''(x)?)

differentiate f(x) to get f'(x)
set f'(x)=0
solve for x
then test for its nature - using either f''(x) or testing either side of the stat.pt. with f'(x)

P!xel · Feb 15, 2009

Thanks

I was playing around with the differentials and found that

When f'(x) = 0, it's a stationary point

When f''(x) = 0, it's a point of inflexion.

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Polynomials and Remainder Theorem (1 Viewer)

P!xel

The lazy one.

gurmies

Drover

kurt.physics

Member

P!xel

The lazy one.

tacogym27101990

Member

P!xel

The lazy one.

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