Hello haohaoxNienie!
Show that p(x) = x³ - 6x² + 11x - 6 is divisible by x - 1.
To show that p(x) = x³ - 6x² + 11x - 6 is divisible by x - 1, we must divide the polynomial by x - 1. If we get the remainder as 0, then we can say that p(x) = x³ - 6x² + 11x - 6 is divisible by x - 1.
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So, let's do the division (steps in the attached picture).
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We got the remainder as 0. So, yes p(x) = x³ - 6x² + 11x - 6 is divisible by x - 1.
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<h3><u>More steps :-</u></h3>
x² - 5x + 6
= <u>(x - 3) (x - 2)</u> [use the splitting the middle term method].
So, the zeroes are,
(x - 3)
x - 3 = 0
<u>x = 3 ⇨ 1</u><u>s</u><u>t</u><u> </u><u>zero.</u>
(x - 2)
x - 2 = 0
<u>x = 2 ⇨ 2nd zero.</u>
Now, use the values of x on the equation. If we get 0 as the solution of the equation then we can prove that p(x) = x³ - 6x² + 11x - 6 is divisible by x - 1.
p(x) = x³ - 6x² + 11x - 6
p(3) = 3³ - 6(3)² + 11 (3) - 6
p(3) = 27 - 54 + 33 - 6
<u>p(3) = 0.</u>
p(x) = x³ - 6x² + 11x - 6
p(2) = 2³ - 6(2)² + 11 (2) - 6
p(2) = 8 - 24 + 22 - 6
<u>p(2) = 0</u>
Since we got the solutions as 0 we can say that p(x) = x³ - 6x² + 11x - 6 is divisible by x - 1.
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Hope it'll help you!
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