The polynomial P(x) of degree 4 has
a root of multiplicity 2 at x = 4
• a root of multiplicity 1 at x = 0 and at x = -1
• It goes through the point (5,21)
Find a formula for P(x).
P(x) =
Question Help: Message instructor

Answer:P(x) = 0.7 [tex]x^{4}[/tex] - 4.9x³ + 5.6x² + 11.2xStep-by-step explanation:given a polynomial with roots x = a and x = b , then the factors are(x - a) and (x - b)If x = a is of multiplicity 2 then factor is (x - a)²the polynomial is then the product of the factorsp(x) = a(x - a)(x - b) ← a is a multipliergive x = 4 is a root with multiplicity 2 then (x - 4)² is the factorx = 0 has factor (x - 0) , that is xx = - 1 has factor (x - (- 1)) , that is (x + 1)the polynomial is then the product of the factorsP(x) = ax(x + 1)(x - 4)² ← expand squared factor using FOIL        = ax(x + 1)(x² - 8x + 16)        = a(x² + x)(x² - 8x + 16) ← distribute        = a([tex]x^{4}[/tex] - 8x³ + 16x² + x³ - 8x² + 16x)       = a([tex]x^{4}[/tex] - 7x³ + 8x² + 16x)to find a substitute (5, 21 ) into P(x)21 = a([tex]5^{4}[/tex] - 7(5)³ + 8(5)² + 16(5))21 = a(625 - 875 + 200 + 80)21 = 30a ( divide both sides by 30 )0.7 = athenP(x) = 0.7([tex]x^{4}[/tex] - 7x³ + 8x² + 16x) ← distribute parenthesisP(x) = 0.7[tex]x^{4}[/tex] - 4.9x³ + 5.6x² + 11.2x