Find the zeroes of the polynomial h(x) = x^3 - 6x^2 + 11x - 6.

Collected in the board: Polynomials

Steven Zheng posted 3 hours ago


To find the zeroes of a polynomial, we set it equal to zero and solve for x. In this case, we can try the Rational Root Theorem, which says that any rational root of h(x) must be of the form p/q, where p is a factor of the constant term (in this case, 6), and q is a factor of the leading coefficient (in this case, 1). So the possible rational roots of h(x) are:

±1, ±2, ±3, ±6

We can try these values in turn and see which ones make h(x) equal to zero. For example, if we plug in x = 1, we get:

h(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 0

So 1 is a zero of h(x). We can use polynomial division or synthetic division to factor h(x) and find the other zeroes:

(x - 1)(x - 2)(x - 3) = 0

So the zeroes of h(x) are x = 1, 2, and 3.

Steven Zheng posted 3 hours ago

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