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αx3 + bx2 + cx + d = 0
In the equation above, α, b, c, and d are constants.
If the equation has roots −1 , −3 , and 5, which of the following is a factor of αx3 + bx2 + cx + d?
In the equation above, α, b, c, and d are constants.
If the equation has roots −1 , −3 , and 5, which of the following is a factor of αx3 + bx2 + cx + d?
- a)x − 1
- b)x + 1
- c)x − 3
- d)x + 5
Correct answer is option 'B'. Can you explain this answer?
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αx3 + bx2+ cx + d = 0In the equation above, α, b, c, and d...
Choice B is correct. In general, a binomial of the form x + f, where f is a constant, is a factor of a polynomial when the remainder of dividing the polynomial by x + f is 0. Let R be the remainder resulting from the division of the polynomial P(x) = αx3 + bx2 + cx + d by x + 1. So the polynomial P(x) can be rewritten as P(x) = (x + 1)q(x) + R, where q(x) is a polynomial of second degree and R is a constant.
Since –1 is a root of the equation P(x) = 0, it follows that P(–1) = 0.
Since P(–1) = 0 and P(–1) = R, it follows that R = 0. This means that x + 1 is a factor of P(x).
Choices A, C, and D are incorrect because none of these choices can be a factor of the polynomial P(x) = αx3 + bx2 + cx + d. For example, if x – 1 were a factor (choice A), then P(x) = (x –1)h(x), for some polynomial function h. It follows that P(1) = (1 – 1)h(1) = 0, so 1 would be another root of the given equation, and thus the given equation would have at least 4 roots. However, a third-degree equation cannot have more than three roots. Therefore, x – 1 cannot be a factor of P(x).
Since –1 is a root of the equation P(x) = 0, it follows that P(–1) = 0.
Since P(–1) = 0 and P(–1) = R, it follows that R = 0. This means that x + 1 is a factor of P(x).
Choices A, C, and D are incorrect because none of these choices can be a factor of the polynomial P(x) = αx3 + bx2 + cx + d. For example, if x – 1 were a factor (choice A), then P(x) = (x –1)h(x), for some polynomial function h. It follows that P(1) = (1 – 1)h(1) = 0, so 1 would be another root of the given equation, and thus the given equation would have at least 4 roots. However, a third-degree equation cannot have more than three roots. Therefore, x – 1 cannot be a factor of P(x).
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αx3 + bx2+ cx + d = 0In the equation above, α, b, c, and d are constants.If the equation has roots −1 , −3 , and 5, which of the following is a factor of αx3 + bx2+ cx + d?a)x − 1b)x + 1c)x − 3d)x + 5Correct answer is option 'B'. Can you explain this answer?
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