Gujarat BoardEnglish MediumSTD 8MATHSFactorization5 Marks
Question
Divide $24\left(x^2 y z+x y^2 z+x y z^2\right)$ by 8xyz using both the methods.
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Answer
$24\left(x^2 y z+x y^2 z+x y z^2\right)$
$= 2 \times 2 \times 2 \times 3 \times [(x \times x \times y \times z) + (x \times y \times y \times z) + (x \times y \times z \times z)]$
$= 2 \times 2 \times 2 \times 3 \times x \times y \times z \times (x + y + z)$ [By taking out the common factor]
$= 8 \times 3 \times xyz \times (x + y + z)$
Now, 2$4\left(x^2 y z+x y^2 z+x y z^2\right) \div 8xyz$
$= \frac {8\times 3 ×xyz \times (x + y + z)}{8\times xyz} = 3 \times (x + y + z) = 3(x + y + z)$
Alternately, $24\left(x^2 y z+x y^2 z+x y z^2\right) \div 8xyz$
$=\frac{24 x^{2} y z}{8 x y z}+\frac{24 x y^{2} z}{8 x y z}+\frac{24 x y z^{2}}{8 x y z}$
$= 3x + 3y + 3z = 3(x + y + z)$
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