Question

Write the general polynomial p(x) if its only zeros are 1,4, and -3, with multiplicities 3,2, and 1, respectively. What is its degree? Find p(x) precisely if p(0)=16.

Answer 1

Answer:

$p(x) = k(x -1)^{3} * (x -4)^{2} * (x +3)$ --- General polynomial

The degree is 6

$p(x) = -\frac{1}{3}(x -1)^{3} * (x -4)^{2} * (x +3)$ --- The polynomial precisely

Step-by-step explanation:

Given

$(a,b,c)= (1,4,-3)$ --- zeros

$(m_1,m_2,m_3) =(3,2,1)$ --- multiplicities

Solving (a): The polynomial

The polynomial is represented as:

$p(x) = (x -a)^{m_1} * (x -b)^{m_2} * (x -c)^{m_3}$

So, we have:

$p(x) = (x -1)^{3} * (x -4)^{2} * (x +3)^{1}$

$p(x) = (x -1)^{3} * (x -4)^{2} * (x +3)$

Include constant k; So, the polynomial becomes

$p(x) = k(x -1)^{3} * (x -4)^{2} * (x +3)$

Solving (b): The degree (d)

To do this, add up the multiplicities

$d = m_1 +m_2 +m_3$

$d = 3+2+1$

$d = 6$

The degree is 6

Solving (c): The actual polynomial

In (a), we have:

$p(x) = k(x -1)^{3} * (x -4)^{2} * (x +3)$

$p(0)=16$ means:

$(x,p(x)) = (0,16)$

This gives:

$16 = k(0 -1)^{3} * (0 -4)^{2} * (0 +3)$

$16 = k(-1)^{3} * (-4)^{2} * (3)$

$16 = k*-48$

Solve for k

$k = \frac{16}{-48}$

$k = -\frac{1}{3}$

So:

$p(x) = k(x -1)^{3} * (x -4)^{2} * (x +3)$

$p(x) = -\frac{1}{3}(x -1)^{3} * (x -4)^{2} * (x +3)$