Divide 39x^3(50x^2-98) by 26x^2(5x+7)
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Gauri Chhatwani 5 years, 4 months ago
- 1 answers
Related Questions
Posted by Savitha Savitha 3 months, 3 weeks ago
- 0 answers
Posted by Akhilesh Patidar 3 months, 3 weeks ago
- 0 answers
Posted by T Prudhvi 4 weeks, 1 day ago
- 0 answers
Posted by Alvin Thomas 3 months, 2 weeks ago
- 0 answers
Posted by Gnani Yogi 3 months, 2 weeks ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 5 years, 4 months ago
39y3 (50y2 – 98) ÷ 26y2(5y + 7)
{tex} = \frac{{39{y^3}(50{y^2} - 98)}}{{26{y^2}(5y + 7)}}{/tex}
{tex} = \frac{{39{y^3} \times 2 \times (25{y^2} - 49)}}{{26{y^2}(5y + 7)}}{/tex}
{tex} = \frac{{39{y^3} \times 2 \times \{ {{(5y)}^2} - {{(7)}^2}\} }}{{26{y^2}(5y + 7)}}{/tex}
{tex} = \frac{{39{y^3} \times 2 \times (5y + 7)(5y - 7)}}{{26{y^2}(5y - 7)}}{/tex}. . . . [Using Identity III
= 3y (5y – 7)
3Thank You