Q. 4 F4.7( 12 Votes )

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Answer :

In the given term


Dividend = 26z3(32z2–18)


Take out the common factor in binomial term


2 × 13 × z × z × z (2 × 2 × 2 × 2 × 2 × z × z – 2 × 3 × 3)


2 × 2 × 13 × z × z × z (2 × 2 × 2 × 2 × z × z – 3 × 3)


52z3(16z2 – 9)


In given expression (16z2 – 9)


Both terms are perfect square


16z2 = 4z × 4z


9 = 3 × 3


(16z2 – 9) Seems to be in identity a2-b2 = (a + b)(a-b)


Where a = 4z and b = 3;


(16z2 – 9) = (4z + 3)(4z – 3)


Hence the factors of (16z2 – 9)are (4z + 3) and (4z – 3)


Divisor = 13z2(4z – 3)


=


=


= 4z(4z + 3)


Hence dividing 26z3(32z2–18) by 13z2(4z – 3) gives out 4z(4z + 3)


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