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