Question

Obtain all the zeroes of the polynmial x⁴ + 6x³ + x² - 24x + 20 if two of its zeroes are 2 and -5

Grade 10, Polynomials.
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Answer 1

Answer:

-2 and - 1

Step-by-step explanation:

bro instead of -5 it should be 5

Answer 2

A root or a zero of a polynomial are the value(s) of X that cause the polynomial to = 0 (or make Y=0). It is an X-intercept. The root is the X-value, and zero is the Y-value. It is not saying that imaginary roots = 0.

x^4 + 6x^3 - x^2 - 6x =0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0We get x = -6

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0We get x = -6CASE 3 : when x^3 - x = 0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0We get x = -6CASE 3 : when x^3 - x = 0x ( x^2 -1) = 0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0We get x = -6CASE 3 : when x^3 - x = 0x ( x^2 -1) = 0Therefore , x^2 - 1 = 0

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0We get x = -6CASE 3 : when x^3 - x = 0x ( x^2 -1) = 0Therefore , x^2 - 1 = 0x^ 2 = 1

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0We get x = -6CASE 3 : when x^3 - x = 0x ( x^2 -1) = 0Therefore , x^2 - 1 = 0x^ 2 = 1Therefore: x= +1 or -1

x^4 + 6x^3 - x^2 - 6x =0x^3(x +6) - x(x + 6) =0(x+6) (x^3 - x ) =0Therefore either ( x+6) = 0 or (x^3 - x) =0CASE 1 : when x+ 6 = 0We get x = -6CASE 3 : when x^3 - x = 0x ( x^2 -1) = 0Therefore , x^2 - 1 = 0x^ 2 = 1Therefore: x= +1 or -1Hence the roots of this equation are (-6,-1, +1) .