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: x=-2, x=3+[tex]\sqrt{2}[/tex], 3-[tex]\sqrt{2}[/tex]Step-by-step explanation:f(x) = x³ - 4x² - 5x + 14-2    |    1    -4     -5     14                 -2      12   -14            1    -6       7     0x³ - 4x² - 5x + 14 = (x+2)(x²-6x+7)x²-6x+7=0x²-6x+9-2=0(x²-6x+9)-2=0x²-6x+9=2x²-3x-3x+9=2x(x-3)-3(x-3)=2(x-3)²=2x-3=[tex]\sqrt{2}[/tex], -[tex]\sqrt{2}[/tex]x=3+[tex]\sqrt{2}[/tex], 3-[tex]\sqrt{2}[/tex]Hence, zeroes are x=-2, x=3+[tex]\sqrt{2}[/tex], 3-[tex]\sqrt{2}[/tex]

Step-by-step explanation:our first "suspicion" for a factoring is an integer factor of the constant term : 14.14 = 2 × 7 or -2 × -7I started to try 2 or -2 and found that -2 is indeed a zero solution. so, the first factors aref(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 0f(x) = (x + 2)(x² - 6x + 7)the zeros for x² - 6x + 7 we get from the general solution for a quadratic equationax² + bx + c = 0x = (-b ± sqrt(b² - 4ac))/(2a)in our casex = (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)