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If ƒ (x) = x³ – 11x² +26x - 16 and x - 1 is a factor of f(x), then find all of

the zeros of f(x) algebraically.

1 Answer

6 votes

If x - 1 is a factor of ƒ(x), then we can write the function as ƒ(x) = (x - 1) * g(x), where g(x) is another polynomial. This means that whenever x - 1 equals zero, the function ƒ(x) will also be equal to zero. Therefore, x - 1 is a zero of the function ƒ(x).

To find the other zeros of ƒ(x), we can use the factor theorem, which states that if a polynomial ƒ(x) has a factor (x - a), then a is a zero of ƒ(x). In this case, we know that x - 1 is a factor of ƒ(x), so 1 is a zero of ƒ(x).

To find the other zeros of ƒ(x), we can divide ƒ(x) by (x - 1) using polynomial long division. This will give us the quotient g(x), which will be a polynomial of degree 2. Since a polynomial of degree 2 can have at most 2 zeros, we know that the other zeros of ƒ(x) must be the zeros of g(x).

To find the zeros of g(x), we can use the quadratic formula, which states that the zeros of a quadratic polynomial ax² + bx + c are given by x = (-b ± √(b² - 4ac)) / (2a). In this case, we can plug in the coefficients of g(x) to find its zeros:

g(x) = x² - 10x + 15

x = (-(-10) ± √((-10)² - 4 * 1 * 15)) / (2 * 1)

x = (10 ± √(100 - 60)) / 2

x = (10 ± √40) / 2

x = (10 ± 2√10) / 2

Therefore, the zeros of ƒ(x) are 1, (10 + 2√10) / 2, and (10 - 2√10) / 2. These are the only values of x for which ƒ(x) equals zero.

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User Drakes
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