1.

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________क्रीडाक्षेत्रम्अगच्छन्।ch-6 Sanskrit class 7 cbse ​

Answer»

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


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