need some helps on polynomial (1 Viewer)

[noobking]

hey guys, could someone please help me with these problems?:wave:


1. if the polynomial 2x^2+4x+4 and a(x+1)^2+b(x+2)^2+c(x+3)^2 are equal for 3 values of x, find a, b and c.

2. the line y=k crosses the curve y=ax^3+bx^2+cx+d four times. find the values of a, b,c,d and k.

3. by factoring the difference F(x)=P(x)-Q(x), describe the intercection between the curves P(x) and Q(x). P(x)=x^4+4x^3-x+5, Q(x)=x^3-3x^2-2x+5

4. P(x) is an odd polynomial of degree 3. it has (x+4) as a factor. and when it's divided by x-3 the remainder is 21. find P(x)

CHEERS:;;

 

[SoulSearcher]

For 1, sub in x = -1, x = -2 and x = -3 to find equations in a, b and c, then solve simultaneously.

2, a = 0, b = 0, c = 0, d = k, I'll get back to you on the reasons why later.

3, the intersection points of the original polynomials become the zeros of the resulting polynomial.

4, You have P(x) = ax³ + bx² + cx + d. P(-4) = 0, so sub in x = -4, P(3) = 21, so sub in x = 3, you have two equations in a, b, c and d. But since P(x) is an odd function, f(-x) = -f(x), therefore you can also substitute in P(4) = 0 and P(-3) = -21 to get 2 more equations in a, b, c and d. Then solve simultaneously. At least I think that's how you do 4, not sure about that method yet.

 

[GaDaMIt]

2. the line y=k crosses the curve y=ax^3+bx^2+cx+d four times. find the values of a, b,c,d and k.

2, a = 0, b = 0, c = 0, d = k, I'll get back to you on the reasons why later.

That would be cause its the zero polynomial, as it has four zeros, but degree 3.

This goes against the rules as a polynomial of degree n has at most n zeroes; in our case it is degree 3, but 4 zeros, thus the zero polynomial

Working:

since y = ax^3 + bx^2 + cx + d

y = k

so

ax^3 + bx^2 + cx + d = k

ax^3 + bx^2 + cx + (d - k) = 0

Since it is the zero polynomial, a=0, b=0, c=0, d=k

a/b/c = 0, d = k