The total cost of (x+1) cameras is ₹ (x^{3}+3 | vedclass

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(D) To find the cost of one camera,we divide the total cost by the number of cameras.Cost of one camera = $\frac{x^{3}+3x^{2}+5x+3}{x+1}$.Using polyno

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

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(D) To find the cost of one camera,we divide the total cost by the number of cameras.Cost of one camera = $\frac{x^{3}+3x^{2}+5x+3}{x+1}$.Using polynomial long division:$1$. Divide $x^{3}$ by $x$ to get $x^{2}$.$2$. Multiply $x^{2}(x+1) = x^{3}+x^{2}$.$3$. Subtract $(x^{3}+3x^{2}+5x+3) - (x^{3}+x^{2}) = 2x^{2}+5x+3$.$4$. Divide $2x^{2}$ by $x$ to get $2x$.$5$. Multiply $2x(x+1) = 2x^{2}+2x$.$6$. Subtract $(2x^{2}+5x+3) - (2x^{2}+2x) = 3x+3$.$7$. Divide $3x$ by $x$ to get $3$.$8$. Multiply $3(x+1) = 3x+3$.$9$. Subtract $(3x+3) - (3x+3) = 0$.The quotient is $x^{2}+2x+3$. Thus,the cost of one camera is ₹ $(x^{2}+2x+3)$.

The total cost of $(x+1)$ cameras is ₹ $(x^{3}+3x^{2}+5x+3)$. Find the cost of one camera.

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