Question
Using factor theorem, factorize the following polynomials: $x^3 + 13x^2 + 32x + 20$
✓
Answer
Let $p(x) = x^3 + 13x^2+ 32x + 20$
The factors of 20 are $\pm1,\pm2,\pm4,\pm5\dots$ By hit and trial method $p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 - 32 + 20 = 33 - 33 = 0 As p(-1)$ is zero,
so $x + 1$ is a factor of this polynomial $p(x).$
Let us find the quotient while dividing $x^3 + 13x^2 + 32x + 20 by (x + 1)$ By long division
We know that Dividend = Divisor $\times $ Quotient + Remainder
$x^3 + 13x^2 + 32x + 20 $
$= (x + 1)(x^2 + 12x + 20) + 0 $
$= (x + 1)(x^2 + 10x + 2x + 20) $
$= (x + 1)[x(x + 10) + 2(x + 10)] $
$= (x + 1)(x + 10)(x + 2) $
$= (x + 1)(x + 2)(x + 10)$
The factors of 20 are $\pm1,\pm2,\pm4,\pm5\dots$ By hit and trial method $p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20 = -1 + 13 - 32 + 20 = 33 - 33 = 0 As p(-1)$ is zero,
so $x + 1$ is a factor of this polynomial $p(x).$
Let us find the quotient while dividing $x^3 + 13x^2 + 32x + 20 by (x + 1)$ By long division
We know that Dividend = Divisor $\times $ Quotient + Remainder
$x^3 + 13x^2 + 32x + 20 $
$= (x + 1)(x^2 + 12x + 20) + 0 $
$= (x + 1)(x^2 + 10x + 2x + 20) $
$= (x + 1)[x(x + 10) + 2(x + 10)] $
$= (x + 1)(x + 10)(x + 2) $
$= (x + 1)(x + 2)(x + 10)$
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