Factorise:x^3-3x^2-9x-5 by without long division.
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Answered by
11
x³ - 3x² - 9x - 5
= x³ + x² - 4x² - 4x - 5x - 5
= x²( x + 1) - 4x ( x + 1) - 5(x + 1)
= (x + 1)(x² - 4x - 5)
= (x + 1)(x² -5x + x - 5)
= (x + 1)(x - 5) (x + 1)
hence, (x +1), (x -5) and (x +1) are the factors of given polynomial
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= x³ + x² - 4x² - 4x - 5x - 5
= x²( x + 1) - 4x ( x + 1) - 5(x + 1)
= (x + 1)(x² - 4x - 5)
= (x + 1)(x² -5x + x - 5)
= (x + 1)(x - 5) (x + 1)
hence, (x +1), (x -5) and (x +1) are the factors of given polynomial
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Answered by
4
Answer:
= ( x - 5) ( x + 1)
Step-by-step explanation:
p(x) = x3 - 3x2 - 9x - 5
Factors of 5 = 1,5,-1,-5
p(1) = 13 - 3 x (1)2 - 9 x (1) - 5
= 1 -3 - 9 - 5
= 1 - 17 = - 16 ≠ 0
p(-1) = (-1)3 x 3 x (-1)2 - 9 x (-1) -5
= -1 - 3 + 9 - 5 = -9 + 9 = 0
p(-1) =0
∴ (x+1) is a factor of p(x)
p(x) ÷ (x+1)
x3 - 3x2 - 9x - 5 ÷ x + 1
= x2 - 4x - 5
x2 - 4x - 5
x2 - 5x + 1x - 5
= x ( x - 5) + 1 ( x - 5)
= ( x - 5) ( x + 1)
(answer)
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