Find coefficient in cubic polynomial

Question

We have $$ p(x) = ax^3 + bx^2 + cx +d$$ where $a, b, c, d$ are complex coefficients. We have to find all posible coefficients for: $$ p(1) = 2$$ $$ p(i) = i$$ $$ p(-1) = 0$$ Unfortunately I dont know how to start... Have I just put values of $ p(x)$ to the polynomial and calculate it? I need solution with parameter and after that I have to find polynomial for $$p(1+i) = 0 $$

Answer 1

Here is an approach that requires only calculation and no solving, using the Lagrange interpolation formula.

Consider the three functions $$ p_1(x) = \frac{2}{(1-i)(1-(-1))}(x-i)(x-(-1)) = \frac{1+i}2(x-i)(x+1)\\ p_2(x) = \frac{i}{(i-1)(i-(-1))}(x-1)(x-(-1)) = \frac i2(x-1)(x+1)\\ p_3(x) = \frac{0}{(-1-1)(-1-i)}(x-1)(x-i) = 0 $$ Note that $p_1(1) = 2$, $p_2(i) = i$ and $p_3(-1) = 0$, while $$ p_1(i) = p_1(-1) = 0\\ p_2(1) = p_2(-1) = 0\\ p_3(1) = p_3(i) = 0 $$ This means that $$ q(x) = p(x) - (p_1(x) + p_2(x) + p_3(x)) $$ has the property that $q(1) = q(i) = q(-1) = 0$. And since the degree of $q$ is (at most) $3$, that means that $q$ must be of the form $a(x-1)(x-i)(x-(-1))$ for some constant $a$. And any such $a$ gives a valid $q$. In other words, $$ p(x) = q(x) + p_1(x) + p_2(x) + p_3(x) $$ gives all possible solutions to what $p$ could possibly be.

Answer 2

$$P(1) =2 \implies a+b+c+d=2 \quad \quad \text {(1.)}$$ $$P(-1) =0 \implies -a+b-c+d=0 \quad \quad \text {(2.)}$$

Subtracting $(2.)$ from $(1.)$ ,

$$2a + 2c =2 \implies a+c=1 \quad \quad \text {(3.)}$$

$$P(i) =i \implies -ia - b + ic+ d=i \quad \quad \text {(4.)}$$

Assuming $a = p \,+\,qi$ , we get $c= r \,- \, qi$ from$ \text {( 4. )}$

Now $$-ia -b+ic+d = i $$

$$ -ip +q -b + ri+q +d =i \implies q=0$$

Hence $a$ and $c$ are real numbers.

Now equating real and imaginary part of $(4.)$ , we get $a=0 ,b=\frac12 ,c=1$ and $d = \frac12$.

So $P(x) = \frac12 x^2 + x + \frac 12.$

Answer 3

$$P(x)=a(x^2-1)(x-i)+px^2+qx+r$$ $$P(1)=p+q+r=2$$ $$P(i)=-p+iq+r=i$$ $$P(-1)=p-q+r=0$$ Solving we get $q=1$ , $p= r=1/2$ $$P(x)=a(x^2-1)(x-i)+(1/2)x^2+x+(1/2)$$

Answer 4

If you set $q(x)=p(x)-(x+1)$ you have that $q(x)$ is a cubic polynomial fulfilling $q(-1)=q(1)=0$ and $q(i)=-1$. The new constraints give $q(x)=(x^2-1)r(x)$ with $r(x)$ being a linear polynomial such that $r(i)=\frac{1}{2}$, i.e. $$ r(x) = k(x-i)+\frac{1}{2}.$$ Working backwards it is simple to find the structure of the coefficients of $q(x)$, then the structure of the coefficients of $p(x)$.