4.
कोष्ठके लिखित-धातुभिः सह 'तुमुन्' प्रत्ययं योजयित्वा रिक्तस्थानानि पूरयन्तु-
(i) अहं प्रश्नं _______इच्छामि।
(ii) शिशुः पादाभ्यां_______इच्छति।
(iii) वयम् अधुना_______गच्छामः।
(iv) सः बालकः चिकित्सकः______
इच्छति।
(v) जनकः देवं________देवालयं गच्छति।
(vi) माताभोजनं _____________
पाकशालायाम् अस्ति।
(vii) ते________क्रीडाक्षेत्रम्
अगच्छन्।
ch-6 Sanskrit class 7 cbse
Answers
Answer:
The first term is, 4a2 its coefficient is 4 .
The middle term is, -13a its coefficient is -13 .
The last term, "the constant", is +10
Step-1 : Multiply the coefficient of the first term by the constant 4 • 10 = 40
Step-2 : Find two factors of 40 whose sum equals the coefficient of the middle term, which is -13 .
-40 + -1 = -41
-20 + -2 = -22
-10 + -4 = -14
-8 + -5 = -13 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -8 and -5
4a2 - 8a - 5a - 10
Step-4 : Add up the first 2 terms, pulling out like factors :
4a • (a-2)
Add up the last 2 terms, pulling out common factors :
5 • (a-2)
Step-5 : Add up the four terms of step 4 :
(4a-5) • (a-2)
Which is the desired factorization
Don't know answer....
Coz
I'm not Expert in Sanskrit...☹️☹️
Really so so so Sorry siso...