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निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए(i) (x2 – 2x)2 – 23(x2 – 2x) + 120(ii) (x + 2y)2 + 5(x + 2y)(2x + y) + 6(2x + y)2 |
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Answer» (i) (x2 – 2x)2 – 23(x2 – 2x) + 120 x2 – 2x = y =y2 – 23y + 120 = y2 – (8 + 15)y + 120 (120 = 8 × 15) = y2 – 8y – 15y + 120 = y(y – 8) – 15(y – 8) = (y – 15)(y – 8) = (x2 – 2x – 15)(x2 – 2x – 8) (15 = 3 × 5 व 8 = 4 × 2) = [x2 – (5 – 3)x – 15][x2 – (4 – 2)x – 8] = [x2 – 5x + 3x – 15][x2 – 4x + 2x – 8] =[x(x – 5) + 3(x – 5)][x(x – 4) + 2(x – 4)] = [(x + 3)(x – 5)][(x – 4)(x + 2)] (ii) (x + 2y)2 + 5(x + 2y)(2x + y) + 6(2x + y)2 माना x + 2y = m तथा 2x + y = n = m2 + 5mn + 6n2 = m2 +(2 + 3)mn +6n2 = m2 + 2mn + 3mn + 6n2 = m(m + 2n) + 3n(m + 2n) = (m + 2n)(m + 3n) यहाँ [x + 2y + 2(2x + y)][x + 2y + 3(2x + y)] =[x + 2y + 4x + 2y][x + 2y + 6x + 3y] = [5x + 4y][7x + 5y] |
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