The area of a rectangle is expressed by the polynomial $A (x) = 6 x^{2} + 17 x + 12 .$ $A(x) = 6x^2+17x+12.$ What is the perimeter of this rectangle?

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Answer 1 · 2018-06-06T11:19:43

$P (x) = 10 x + 14$ $P(x) = 10x+14$

The area of a rectangle is found from $A = l \times b$ $A = l xx b$

We therefore need to find the factors of the polynomial.

$A (x) = 6 x^{2} + 17 x + 12$ $A(x) = 6x^2+17x+12$

$A (x) = (3 x + 4) (2 x + 3)$ $A(x) = (3x+4)(2x+3)$

We cannot get numerical values for the length and breadth, but we have found them in terms of $x$ $x$ .

$l = (3 x + 4) and b = (2 x + 3)$ $l = (3x+4) and b= (2x+3)$

$P = 2 l + 2 b$ $P = 2l + 2b$

$P (x) = 2 (3 x + 4) + 2 (2 x + 3)$ $P(x) = 2(3x+4) +2(2x+3)$

$P (x) = 6 x + 8 + 4 x + 6$ $P(x) = 6x+8+4x+6$

$P (x) = 10 x + 14$ $P(x) = 10x+14$

The area of a rectangle is expressed by the polynomial A(x) = 6x^2+17x+12.A(x)=6x2+17x+12.A(x) = 6x^2+17x+12. What is the perimeter of this rectangle?