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The perimeter of a rectangle is \( 16 x^{3}-6 x^{2}+12 x+4 \). If one of its sides is \( 8 x^{2}+3 x \), find the other side. |
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Answer» The perimeter of rectangle = 16x³-6x²+12x+4 We know that , perimeter of rectangle =2(I + b) Given that one of it's side is 8x²+3x Let, l =8x²+3x \(\implies\)2(l +b) = 16x³-6x²+12x+4 \(\implies\) b = (8x³-3x²+6x+2)-(8x²+3x) \(\implies\) b = 8x³-11x²+3x+2 The other side = 8x³-11x²+3x+2 Perimeter of rectangle=16x3 − 6x2 + 12x + 4 2(l + b) = 16x3 − 6x2 + 12x + 4 l + b = 8x3 − 3x2 + 6x + 2 b = (8x3 − 3x2 + 6x + 2) − (8x2 + 3x) = 8x3 − 3x2 + 6x + 2 − 8x2 − 3x = 8x3 − 11x2 + 3x + 2 |
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