square 1 The polynomis of degree 4, P(z) has a not of m z=0 and z=-1 . It goes through the point 6. Find a formula for P(z) P(z)= square

To find the polynomial \( P(z) \) of degree 4 with the given roots and point, we can use the fact that if \( z = 0 \) and \( z = -1 \) are roots, then \( z \) and \( z + 1 \) are factors of the polynomial. Additionally, since \( z = 7 \) is a root, \( z - 7 \) is also a factor.<br /><br />Given that the polynomial passes through the point (6, 0), we can use this information to determine the constant factor.<br /><br />The polynomial can be written as:<br />\[ P(z) = k \cdot z \cdot (z + 1) \cdot (z - 7) \]<br /><br />To find the constant \( k \), we use the fact that the polynomial passes through the point (6, 0). This means that when \( z = 6 \), \( P(6) = 0 \).<br /><br />Substituting \( z = 6 \) into the polynomial equation, we get:<br />\[ 0 = k \cdot 6 \cdot (6 + 1) \cdot (6 - 7) \]<br /><br />Simplifying the equation, we have:<br />\[ 0 = k \cdot 6 \cdot 7 \cdot (-1) \]<br />\[ 0 = -42k \]<br /><br />Since the equation is true for any value of \( k \), we can choose \( k = 1 \) for simplicity.<br /><br />Therefore, the polynomial \( P(z) \) is:<br />\[ P(z) = z \cdot (z + 1) \cdot (z - 7) \]<br /><br />Expanding the factors, we get:<br />\[ P(z) = z(z^2 - 6z - 7) \]<br />\[ P(z) = z^3 - 6z^2 - 7z \]<br /><br />So, the final answer is:<br />\[ P(z) = z^3 - 6z^2 - 7z \]