Show that (x-2), (x+3) and (x-4) are factors of x³-3x²-10x+24.
Answers
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Given : (x - 2), (x + 3) and (x - 4) are factors of x³ - 3x² - 10x + 24.
Let p(x) = x³ - 3x² - 10x + 24 be the given polynomial. In order to prove that (x - 2), (x + 3) and (x - 4) are factors of p(x) , it is sufficient to show that p(2) , p(-3) and p(4) are equal to zero.
Now, p(x) = x³ - 3x² - 10x + 24
p(2) = (2)³ – 3(2)² – 10 x 2 + 24
= 8 – 12 – 20 + 24
= 32 – 32
= 0
and,
p(- 3) = (-3)³ - 3(-3)² – 10 (-3) + 24
= - 27 - 27 + 30 + 24
= - 54 + 54
= 0
And,
p(4) = (4)³ – 3(4)² – 10 x 4 + 24
= 64 - 48 - 40 + 24
= 88 – 88
= 0
Hence (x - 2), (x + 3) and (x - 4) are the factors of the given polynomial.
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