1.

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.

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, ​​​=8x²+3x

\(\implies\)2(+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=16x− 6x+ 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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