Question

Find a polynomial function of degree 3 such that f(0)=17 and the square root of f(x) are 0,5 and 8

Answer 1

Answer: Okay, lets explain.

Step-by-step explanation:Since 0, 5 & 8 are given as the zeros of the required 3rd degree polynomial f(x), therefore, one may take it as ; f(x) =k (x-0)(x-5)(x-8)

= k(x³−13x²+40) …. .. .(1) . Since f(10) = 17 (given), it implies 17 = k(1000 -1300 + 400) = 100 ==> k = 17/100 = 0.17 . Put this value of k in eq(1) and get the required polynomial.

Answer 2

f(x)=x^3 - 13x^2 + 40x + 17

Explanation:

x-0=0 x-5=0 x-8=0

x=0 x=5 x=8

y=x(x-5)(x-8)+b ==> b is what's going to be used to find the equation so that

y=17 when x=0

17=0(0-5)(0-8)+b ==> plugin 0 for x and 17 for y

17=0*(-5)*(-8)+b ==> simplify

17=0+b ==> anything multiplied by 0 is 0.

b=17

Hence, the equation is:

y=x(x-5)(x-8)+17 ==> expand this equation

y=x*(x-5)(x-8) + 17

y=x*(x(x - 8) - 5(x - 8)) + 17 ==> distribute x-8 to x and -5

y=x*(x*x - 8x - 5x - (8)(-5)) + 17 ==> distribution property

y=x*(x^2 - 13x - (-40)) + 17 ==> simplify

y=x*(x^2 - 13x + 40) + 17 ==> subtracting a negative number is equivalent to

adding a positive number

y=x^3 - 13x^2 + 40x + 17 ==> multiply x with x^2, 13x, and 40 using the

distribution property.

Answer: f(x)=x^3 - 13x^2 + 40x + 17