Question

Which of the following represents the zeros of f(x) = 6x3 + 25x2 − 24x + 5?

Answer 1

The given polynomial is

`f(x) = 6x^3 + 25x^2 - 24x + 5`

By rational root theorem

For a cubic polynomial of the type :

a x³+ b x² + c x + d=0

→ x³ +
`(bx^2)/(a) +  (c x)/(a)+(d)/(a)=0`

The possible roots of this polynomial is ,
`\pm 1, \pm d, \pm (1)/(a) , \pm(d)/(a)`.

The polynomial f(x) can be written as:
`x^3 +(25x^2)/(6)-4 x+(5)/(6)`

The possible roots or Zeroes of the polynomial are:

`\pm 1,\pm 3, \pm 5,\pm (1)/(2), \pm (1)/(3), \pm (5)/(2)\pm (5)/(3),\pm (5)/(6)`

The values of x for which f(x) =0 , are roots or zeroes of the polynomial.

f(1)= 6 +25 -24+5= 12

f(-1)= -6 +25 +24 +5=48

f(3)=162 +225-72+5≠ 0

f(-3)=-162+225+72+5≠0

`f((-1)/(2)) =(-6)/(8)+(25)/(4)+12+5=(93)/(4)- (6)/(8)\\eq 0`

`f((1)/(2)) =(6)/(8)+(25)/(4)-12+5= 7-7=0`

`f((1)/(3))=(6)/(27)+(25)/(9)-8+5=0 \\\\f((-1)/(3))=(-6)/(27)+(25)/(9)+8+5 \\eq 0`

`f(5)=750 +625-120+5\\eq 0\\ f(-5)= -750+625 +120+5=0\\f((5)/(2))=(750)/(8)+(625)/(4)-60+5\\eq 0\\f((-5)/(2))=(-750)/(8)+(625)/(4)+60+5\\eq 0\\\\`

`f((5)/(3))= (750)/(27)+(625)/(9)-40+5\\eq 0\\\\\\f((-5)/(3))= (-750)/(27)+(625)/(9)+40+5\\eq 0\\\\\\f((5)/(6))= (125)/(36)+(625)/(36)-20+5\\eq 0 \\\\\\\\f((-5)/(6))= (-125)/(36)+(625)/(36)+20+5\\eq 0`

So , zeroes of f(x) are -5,
`(1)/(2)`,
`(1)/(3)`.

Answer 2

The zeros of f(x) are {-5, 1/3, 1/2}.