Question

Find all zeros of f(x) = x³ - 4x² - 5x + 14. Enter the zeros separated by commas. Enter exact values,

using radicals and/or fractions if necessary, not decimal approximations.
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Answer 1

Explanation:

our first "suspicion" for a factoring is an integer factor of the constant term : 14.

14 = 2 × 7 or -2 × -7

I started to try 2 or -2 and found that -2 is indeed a zero solution.

so, the first factors are

f(x) = (x + 2)(x² + ...)

let's divide (x³ - 4x² - 5x + 14) by (x + 2) to get the x² term.

x³ - 4x² - 5x + 14 ÷ x + 2 = x² - 6x + 7

- x³ + 2x²

--------------

0 -6x² - 5x

- -6x² - 12x

----------------

0 7x + 14

- 7x + 14

------------

0 0

f(x) = (x + 2)(x² - 6x + 7)

the zeros for x² - 6x + 7 we get from the general solution for a quadratic equation

ax² + bx + c = 0

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = (6 ± sqrt((-6)² - 4×1×7))/(2×1) =

= (6 ± sqrt(36 - 28))/2 = (6 ± sqrt(8))/2 =

= 3 ± sqrt(8/4) = 3 ± sqrt(2)

x1 = 3 + sqrt(2)

x2 = 3 - sqrt(2)

so, the zeros are

-2, 3 - sqrt(2), 3 + sqrt(2)

Answer 2

Answer: x=-2, x=3+
`√(2)`, 3-
`√(2)`

Explanation:

f(x) = x³ - 4x² - 5x + 14

-2 | 1 -4 -5 14

-2 12 -14

1 -6 7 0

x³ - 4x² - 5x + 14 = (x+2)(x²-6x+7)

x²-6x+7=0

x²-6x+9-2=0

(x²-6x+9)-2=0

x²-6x+9=2

x²-3x-3x+9=2

x(x-3)-3(x-3)=2

(x-3)²=2

x-3=
`√(2)`, -
`√(2)`

x=3+
`√(2)`, 3-
`√(2)`

Hence, zeroes are x=-2, x=3+
`√(2)`, 3-
`√(2)`