3U Polynomials Questions (1 Viewer)

[x.Exhaust.x]

1. Consider the cubic expression p[x] = ax^3 + bx^2 + cx + d

a) Given that p[x] is an odd function, evaluate b and d


f(-x) = -f(x)? Then?
b) It is now given that p[x] is monic. What conditions must exist on c for 3 real roots?

2. Factorise f(x)=x^6-x^5-17x^4+5x^3+64x^2-4x-48

So far I've got factors of 1, 2, -2, -3, 4 which all amount to 0 (remainder theorem). Now my mind is blank as to what to continue on with.

3. Consiver the cubic polynomial p(x) = ax^3 + bx^2 + cx + d

a) Show that p(x)-p(alpha)= (x-alpha).q(x) for some q(x)

b) Deduce the remainder theorem for a cubic polynomial.

Thanks in advance.

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[namburger]

1 a) Evaluate?? There are like no values but i came up with the expression bx^2 + d = 0

2. with the factors 1, 2, -2, -3, 4, you have deduced 5 of the 6 factors. The last factor has to be -1 as the product of roots is -48

Answer: (x-1)(x-2)(x+2)(x+3)(x-4)(x+1)

3. a) ill call alpha "z"

p(x)-p(z) = a (x^3-z^3) + b (x^2-z^2) + c (x-z)
= a (x-z)(x^2+xz+z^2) + b (x-z)(x+z) + c (x-z)
= (x-z) [ a(x^2+xz+z^2) + b(x+z) + c] Through factorisation
= RHS

b)
p(x)-p(alpha)= (x-alpha).q(x)
p(x) = (x-alpha).q(x) + p(alpha)

Therefore the remainder when a polynomial is divided by (x-alpha) is p(alpha)

 

[Trebla]

For a), if the cubic polynomial is odd, then it passes through the origin. Therefore d = 0, and from simplifying P(-x) = - P(x), you end up with bx² + d = 0. Since d = 0, then b = 0 as well for all real x.

Now use this result for b): which leaves us with
P(x) = ax³ + cx
Since P(x) is monic a = 1, hence
P(x) = x³ + cx
= x(x² + c)
Now for P(x) already has one real root x = 0. For P(x) to have three real roots the quadratic factor term (x² + c) must have TWO real solutions. The only way that can happen is if c is non-positive, because if c is positive, then the quadratic has no real solutions. Therefore the condition is c ≤ 0.

NB: The question did not specify if the 3 real roots had to be distinct, so the c = 0 case is possible because three real roots exist but they are all equal to 0 (i.e. a simple degenerate y = x³ curve)

 

[12o9]

 

[namburger]

hmm forgot odd functions past through 0,0. Thanks

and im quite sure the answer for 1.b) is c < 0

 

[bubblesss]

1. Consider the cubic expression p[x] = ax^3 + bx^2 + cx + d

a) Given that p[x] is an odd function, evaluate b and d


f(-x) = -f(x)? Then?

a(-x)^3+bx^2 + -cx +d = p (x)
-ax^3+bx^2-cx+d=-ax^3-bx^2-cx-d
2bx^2+2d=0
bx^2+d=0
bx^2=-d
x^2=-d/b
x=sqroot of -d/b

 

[x.Exhaust.x]

f(-x) = -f(x)? Then?

a(-x)^3+bx^2 + -cx +d = p (x)
-ax^3+bx^2-cx+d=-ax^3-bx^2-cx-d
2bx^2+2d=0
bx^2+d=0
bx^2=-d
x^2=-d/b
x=sqroot of -d/b

Didn't need x. But thanks for trying .

And I don't understand how you get bx^2 + d = 0, hence d=0 and b=0.

My working:

-P(x) = -ax^3-bx-cx-d
P(-x) = -ax^2+bx^2-cx+d

Then what?

 

[12o9]

Didn't need x. But thanks for trying .

And I don't understand how you get bx^2 + d = 0, hence d=0 and b=0.

My working:

-P(x) = -ax^3-bx-cx-d
P(-x) = -ax^2+bx^2-cx+d

Then what?

-P(x)=P(-x)
then -b=b and -d=d
therefore, b=0, d=0

 

[x.Exhaust.x]

-P(x)=P(-x)
then -b=b and -d=d
therefore, b=0, d=0

Ah right thanks