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A cubic with roots of 3,0, and -2 that passed through (-1,6)

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Answer:

Therefore, q(x) must be a constant polynomial, and the cubic polynomial that has roots of 3, 0, and -2, and passes through the point (-1, 6) is:f(x) = (x-3)(x-0)(x+2)(0) + 6

= (x-3)(x)(x+2) + 6

= x^3 - x^2 - 5x + 6

Explanation:

If a cubic polynomial has roots of 3, 0, and -2, and it passes through the point (-1, 6), then we can write the polynomial in the form:f(x) = (x-3)(x-0)(x+2)q(x) + 6where q(x) is a polynomial of degree 0.To find q(x), we can plug in x = -1 to the equation above and solve for q(-1). This gives us:f(-1) = (-1-3)(-1-0)(-1+2)q(-1) + 6

= (-4)(-1)(1)q(-1) + 6

= 4q(-1) + 6

= 6Solving for q(-1) gives us q(-1) = (6-6)/4 = 0.

User Zach Waugh
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