Question
Using factor theorem, factorize the following polynomials:
x3 - 3x2 - 9x - 5
x3 - 3x2 - 9x - 5
✓
Answer
Let p(x) = x3 - 3x2 - 9x - 5
The factors of 5 are $\pm1,\pm5.$
By hit and trial method
p(-1) = (-1)3 - 3(-1)2 - 9(-1) - 5
= -1 - 3 + 9 - 5 = 0
So x + 1 is a factor of this polynomial p(x).
Let us find the quotient while dividing x3 - 3x2 - 9x - 5 by x + 1
By long division
Now,
Dividend = Divisor × Quotient + Remainder
$\therefore$ x3 - 3x2 - 9x - 5 = (x + 1)(x2 - 4x - 5) + 0
= (x + 1)(x2 - 5x + x - 5)
= (x + 1)[x(x - 5) + 1(x - 5)]
= (x + 1)(x - 5)(x + 1)
= (x - 5)(x + 1)(x + 1)
The factors of 5 are $\pm1,\pm5.$
By hit and trial method
p(-1) = (-1)3 - 3(-1)2 - 9(-1) - 5
= -1 - 3 + 9 - 5 = 0
So x + 1 is a factor of this polynomial p(x).
Let us find the quotient while dividing x3 - 3x2 - 9x - 5 by x + 1
By long division
Now,
Dividend = Divisor × Quotient + Remainder
$\therefore$ x3 - 3x2 - 9x - 5 = (x + 1)(x2 - 4x - 5) + 0
= (x + 1)(x2 - 5x + x - 5)
= (x + 1)[x(x - 5) + 1(x - 5)]
= (x + 1)(x - 5)(x + 1)
= (x - 5)(x + 1)(x + 1)
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