Find the zeros of the function f(x)=x^3-7x^2+6x

f(x)=x^3-7x^2+6x

The zeros of the function

x^3-7x^2+6x = 0

x(x^2 -7x +6) = 0

x(x - 6)(x - 1) = 0

x = 0

x - 6 = 0; x = 6

x - 1 = 0; x = 1

Answer

x = 0, 1 , 6

answer: bea's uncle is 25 years old.

step-by-step explanation:

given the equation , we can say that the variable "x" represents the age of bea's uncle. then to find his age, we need to solve for "x":

we must add 15 to both sides of the equation:

now we need to divide both sides of the equation by 3:

therefore, we can conclude that bea's uncle is 25 years old.

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)