The polynomial of degree \( 4, P(x) \) has a root of multiplicity 2 at \( x=2 \) and roots of multiplicity 1 gat \( x=0 \) and \( x=-2 \). It goes through the point \( (5,189) \). Find a formula for \( P(x) \). \( P(x)= \)

To find the formula for the polynomial \( P(x) \) of degree 4 with the given roots and passing through the point \( (5,189) \), we can use the factored form of the polynomial and then apply the remainder theorem to find the constant term. Given: - The polynomial \( P(x) \) has a root of multiplicity 2 at \( x=2 \). - The polynomial has roots of multiplicity 1 at \( x=0 \) and \( x=-2 \). - The polynomial passes through the point \( (5,189) \). First, we can express the polynomial in its factored form based on the given roots: \[ P(x) = a(x-2)^2(x-0)(x-(-2)) \] where \( a \) is a constant to be determined. Next, we can expand the factored form of the polynomial to get the general form of \( P(x) \): \[ P(x) = a(x-2)^2(x)(x+2) \] Now, we can use the point \( (5,189) \) to find the value of \( a \) by substituting \( x=5 \) and \( P(x)=189 \) into the polynomial: \[ 189 = a(5-2)^2(5)(5+2) \] Solving for \( a \) will give us the formula for the polynomial \( P(x) \). Solve the equation \( 189 = a(5-2)^2(5)(5+2) \). Solve the equation by following steps: - step0: Solve for \(a\): \(189=a\left(5-2\right)^{2}\times 5\left(5+2\right)\) - step1: Simplify: \(189=315a\) - step2: Swap the sides: \(315a=189\) - step3: Divide both sides: \(\frac{315a}{315}=\frac{189}{315}\) - step4: Divide the numbers: \(a=\frac{3}{5}\) The value of \( a \) is \( a = \frac{3}{5} \) or \( a = 0.6 \). Now, we can substitute the value of \( a \) back into the factored form of the polynomial to find the formula for \( P(x) \): \[ P(x) = \frac{3}{5}(x-2)^2(x)(x+2) \] Therefore, the formula for the polynomial \( P(x) \) is: \[ P(x) = \frac{3}{5}(x-2)^2(x)(x+2) \]