Question
Using factor theorem, factorize the following polynomials: $x^3 - 3x^2 - 9x - 5$
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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 × 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 × 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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In the given figure, $L$ and $M$ are the mid$-$points of $AB$ and $BC$ respectively.
$i.$ If $AB = BC$, prove that $AL = MC$.
$ii.$ If $BL = BM$, prove that $AB = BC$.
Hint:
$i.\text{AB}=\text{BC}$
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$ii.\text{BL}=\text{BM}$
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→$i.$ If $AB = BC$, prove that $AL = MC$.
$ii.$ If $BL = BM$, prove that $AB = BC$.
Hint:
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$\Rightarrow\frac{1}{2}\text{AB}=\frac{1}{2}\text{BC}$
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$ii.\text{BL}=\text{BM}$
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$\Rightarrow\text{AB}=\text{BC}.$
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