Question

find. question no. 13. and. 14

Answer 1

$(2 + \sqrt{3} ) \: and \: (2 - \sqrt{3} ) \: are \: two \: zeres$

$hence \: (x - 2 - \sqrt{3} )and(x - 2 + \sqrt{3} )are \: factors$

$(x - 2) {}^{2} - ( \sqrt{3} ) {}^{2}$

$x {}^{2} + 4 - 4x - 3$

$x {}^{2} - 4x + 1$

$(2x {}^{4} - 9x {}^{3} + 5x {}^{2} + 3x - 1) \div (x {}^{2} - 4x + 1)$

On dividing u will get

$2x {}^{2} - x - 1$

$2x {}^{2} - 2x + x - 1$

$2x(x - 1) + 1(x - 1)$

$(2x + 1)(x - 1)$

Hence other two zeroes are -1/2 and 1

Here is 13th one

Factore of 96 = 2×2×2×2×2×3

Factors of 404 = 2×2×101

HCF = 2×2 = 4

LCM = 2×2×2×2×2×3×101=9696

HCF × LCM = 404 × 96

4×9696 = 404 × 96

38784 = 38784

Thanks