Factorization of x3−3x2−9x−5 will be ____.

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Detailed Solution

We are given a quadratic equation x3−3x2−9x−5.
Let f(x) = x3−3x2−9x−5
According to the Factor Theorem, k is a zero of f(x) if and only if (x−k) is a factor of f(x).
Substituting x = 1 in f(x), we have
⇒ f(1) = 13−3(1)2−9(1)−5 = −16 ≠ 0
So, x−1 is not a factor of f(x) .
Substituting x = −1 in f(x), we have
⇒f(−1) = (−1)3−3(−1)2−9(−1)−5
Applying the exponent, we get
⇒f(−1) = −1−3+9−5
Adding and subtracting the terms, we get
⇒f(−1) = 9−9 = 0
So, is a factor of f(x).
Now, by using synthetic division, we have So, Divisor = x2−4x−5 and Remainder = 0.
Now by using the formula Dividend = Quotient × Divisor + Remainder, we can write x3−3x2−9x−5 = (x+1)⋅(x2−4x−5)+0
⇒x3−3x2−9x−5 = (x+1)⋅(x2−4x−5)
By factoring, we get
⇒x3−3x2−9x−5 = (x+1)⋅(x2+x−5x−5)
Now taking common factors, we get
⇒x3−3x2−9x−5 = (x+1)⋅(x(x+1)−5(x+1))
Factoring out common terms, we get
⇒x3−3x2−9x−5 = (x+1)(x+1)(x−5)
⇒x3−3x2−9x−5 = (x+1)2(x−5)
Thus the factors are (x+1)2, (x−5)
x3−3x2−9x−5 = (x+1)2(x−5)

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