Polynomials Question (1 Viewer)

cutemouse · Aug 1, 2009

$\text{The polynomial P(x) is given by}\ P(x)=x^3-x^2-8x+12\\(i)\ \text{Factorise P(x) completely given that P(x)=0 has a repeated root}\\(ii)\text{The polynomial Q(x) has the from Q(x)=P(x)(x+a) with P(x) as given above}\\ \text{and where the constant a is chosen so that}\ Q(x)\ge 0\for\ all\ real\ values\ of\ x.\\ \text{Find the possible values of a}$

I can do part (i)... Just need help with (ii). Thanks =D

Trebla · Aug 1, 2009

P(x) = (x - 2)²(x+3)
=> Q(x) = (x - 2)²(x+3)(x + a) ≥ 0

For this to hold true we need (x+3)(x + a) ≥ 0 as (x - 2)² ≥ 0 already for any real x.
It should be obvious now that a = 3 is a solution, because (x + 3)² ≥ 0 for all real x.

gurmies · Aug 1, 2009

P(x) = x^3-x^2-8x+12 = (x+3)(x-2)^2

Q(x) = P(x)(x+a) = (x+a)(x+3)(x-2)^2

Since (x-2)^2 is positive for all x, we need to find (x+a)(x+3) >= 0 for all x. I think this is only possible if a = 3, making Q(x) = (x+3)^2(x+2)^2

EDIT: Damn, beaten.

If it didn't occur to you that it can only be one answer or something, here's another approach:

$(x+3)(x+a)=x^{2}+ax+3x+3a \\\\ \frac{d}{dx}(x^{2}+ax+3x+3a)=2x+a+3 \\\\ The \,\, stationary \,\, point \,\, will \,\, be \,\, a \,\, minimum \,\, due \,\, to \,\, the \,\, positive \,\, leading \,\, coefficient. \\\\ When \,\, \frac{d}{dx}(x^{2}+ax+3x+3a)=0 \Rightarrow x=\frac{-(a+3)}{2} \\\\ We \,\, need \,\, to \,\, solve: \,\,\left \{ \frac{-(a+3)}{2}+3 \right \}\left \{ \frac{-(a+3)}{2}+a \right \} \geq 0 \\\\ After \,\, expanding \,\, and \,\, collecting \,\, like \,\, terms,\,\, a^{2}-6a+9\leq 0 \Rightarrow (a-3)^{2}\leq 0 \\\\ This \,\, is \,\, satisfied \,\ only \,\, by \,\, a = 3$

untouchablecuz · Aug 1, 2009

gurmies said:
P(x) = x^3-x^2-8x+12 = (x+3)(x-2)^2

Q(x) = P(x)(x+a) = (x+a)(x+3)(x-2)^2

Since (x-2)^2 is positive for all x, we need to find (x+a)(x+3) >= 0 for all x. I think this is only possible if a = 3, making Q(x) = (x+3)^2(x+2)^2

EDIT: Damn, beaten.

If it didn't occur to you that it can only be one answer or something, here's another approach:

$(x+3)(x+a)=x^{2}+ax+3x+3a \\\\ \frac{d}{dx}(x^{2}+ax+3x+3a)=2x+a+3 \\\\ The \,\, stationary \,\, point \,\, will \,\, be \,\, a \,\, minimum \,\, due \,\, to \,\, the \,\, positive \,\, leading \,\, coefficient. \\\\ When \,\, \frac{d}{dx}(x^{2}+ax+3x+3a)=0 \Rightarrow x=\frac{-(a+3)}{2} \\\\ We \,\, need \,\, to \,\, solve: \,\,\left \{ \frac{-(a+3)}{2}+3 \right \}\left \{ \frac{-(a+3)}{2}+a \right \} \geq 0 \\\\ After \,\, expanding \,\, and \,\, collecting \,\, like \,\, terms,\,\, a^{2}-6a+9\leq 0 \Rightarrow (a-3)^{2}\leq 0 \\\\ This \,\, is \,\, satisfied \,\ only \,\, by \,\, a = 3$

for the sake of it, ill add my solution aswell (if u can't do it by inspection)

Let f(x) = (x+a)(x+3) = x^2 + 3x + ax + 3a = x^2 + x(a+3) + 3a

completing the square,

f(x) = (x + a/2 + 3/2)^2 + [3a - (a/2 + 3/2)^2] =

we find a such that [3a - (a/2 + 3/2)^2] >= 0, so that f(x) >= 0 (i.e. positive definate parabola)

simplifying, (a - 3)^2 <= 0

so a = 3

Polynomials Question (1 Viewer)

cutemouse

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Trebla

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gurmies

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