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Divide 39x^3(50x^2-98) by 26x^2(5x+7)

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Divide 39x^3(50x^2-98) by 26x^2(5x+7)

Posted by Gauri Chhatwani 5 years, 10 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 5 years, 10 months ago

39y³ (50y²– 98) ÷ 26y²(5y + 7)
$= \frac{{39{y^3}(50{y^2} - 98)}}{{26{y^2}(5y + 7)}}$
$= \frac{{39{y^3} \times 2 \times (25{y^2} - 49)}}{{26{y^2}(5y + 7)}}$
$= \frac{{39{y^3} \times 2 \times \{ {{(5y)}^2} - {{(7)}^2}\} }}{{26{y^2}(5y + 7)}}$
$= \frac{{39{y^3} \times 2 \times (5y + 7)(5y - 7)}}{{26{y^2}(5y - 7)}}$ . . . . [Using Identity III
= 3y (5y – 7)

3Thank You

ANSWER

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