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Show that (x – 3) is a factor of x3 + 9x2 – x – 105. |
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Answer» Let p(x) = x3 + 9x2 – x – 105 By factor theorem, x – 3 is a factor of p(x), if p(3) = 0 p(3) = 33 + 9(3)3 – 3 – 105 = 27 + 81 – 3 – 105 = 108 – 108 p(3) = 0 To find the zero of x - 3: Put x - 3 = 0 We get x = 3 Therefore, x – 3 is a factor of x3 + 9x2 – x – 105. |
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