Question

Which of the following represents the zeros of f(x) = x3 − 12x2 + 41x − 42?

7, −3, 2
7, −3, −2
7, 3, 2
7, 3, −2

Answer 1

Answer: Zeroes are,

7, 3, 2

Explanation:

Here, the given cubic equation,

`f(x) = x^3 - 12x^2 + 41x - 42`

Since, at x = 7,

`f(7)=(7)^3-12* 7^2+41* 7 - 42 = 343 - 12* 49 +287 - 42 = 343 - 588 + 245=0`

Thus, 7 is one of the zeroes of f(x),

⇒ x - 7 is a factor of f(x),

By the long division method ( shown below ),

We found that,

`x^3 - 12x^2 + 41x - 42=(x-7)(x^2-5x+6)`

`=(x-7)(x^2-3x-2x+6)` ( By middle term splitting )

`=(x-7)(x(x-3)-2(x-3))`

`=(x-7)(x-3)(x-2)`

For finding the zeroes, f(x) = 0,

`(x-7)(x-3)(x-2)=0`

⇒ x -7 =0 or x-3 =0 or x-2 =0

⇒ x = 7 or 3 or 2

Answer 2

Given the polynomial function
`f(x)=x^3-12x^2+41x-42.`

The integer zeros should be divisors of free term -42.

The divisors of -42 are:

`\pm 1, \pm 2, \pm 3, \pm 6, \pm 7, \pm 14, \pm 21, \pm 42.`

If some of these number is zero of f(x), then f takes 0 value at this point.

Check them:

`f(1)=1^3-12\cdot 1^2 + 41\cdot 1-42=1-12+41-42=-12\\eq 0;`

`f(-1)=(-1)^3-12\cdot (-1)^2 + 41\cdot (-1)-42=-1-12-41-42=-96\\eq 0;`

`f(2)=2^3-12\cdot 2^2 + 41\cdot 2-42=8-48+82-42=0;`

`f(-2)=(-2)^3-12\cdot (-2)^2 + 41\cdot (-2)-42=-8-48-82-42=-180\\eq 0;`

`f(3)=3^3-12\cdot 3^2 + 41\cdot 3-42=27-108+123-42=0;`

`f(-3)=(-3)^3-12\cdot (-3)^2 + 41\cdot (-3)-42=-27-108-123-42=-300\\eq 0;`

`f(6)=6^3-12\cdot 6^2 + 41\cdot 6-42=216-432+246-42=-12\\eq 0;`

`f(-6)=(-6)^3-12\cdot (-6)^2 + 41\cdot (-6)-42=-216-432-246-42=-936\\eq 0;`

`f(7)=7^3-12\cdot 7^2 + 41\cdot 7-42=343-588+287-42=0.`

Since the third degree polynomial function may have only 3 zeros, then you can end this process and state that zeros are 2, 3 and 7, because f(2)=0, f(3)=0 and f(7)=0.

Answer: correct choice is C