Sometimes, in mathematical problems we are required to find certain lengths, areas and volumes in terms of a given variable. We will illustrate the idea with two examples:
Suppose we know that the length of a certain rectangle is $6$6 $cm$cm more than twice its width. Can we find its area and perimeter?
On this information alone, we can't. But what we can do is formulate an expression - specifically a polynomial expression - for the perimeter and area in terms of the width of the rectangle.
Suppose we call the width of the rectangle $x$x. Then this means that the length could be expressed as $2x+6$2x+6.
The perimeter $P$P, in terms of $x$x, becomes the linear polynomial $P=2\left[x+\left(2x+6\right)\right]=6x+12$P=2[x+(2x+6)]=6x+12 and the area $A$A, in terms of $x$x, becomes the quadratic polynomial $A=x\left(2x+6\right)=2x^2+6x$A=x(2x+6)=2x2+6x.
These are variable quantities, and remain so until further information is provided.
For example, suppose we learn at a later time that the width is $3$3 $cm$cm. Then we immediately know that:
$P$P | $=$= | $6x+12$6x+12 |
$=$= | $6\left(3\right)+12$6(3)+12 | |
$=$= | $30cm$30cm | |
and
$A$A | $=$= | $2x^2+6x$2x2+6x |
$=$= | $2\left(3\right)^2+6\left(3\right)$2(3)2+6(3) | |
$=$= | $36$36 $cm^2$cm2 | |
Alternatively, perhaps we learn that the numerical values of the area and the perimeter are equal. Then, with the variable quantities known we can solve for $x$x in the equation $2x^2+6x=6x+12$2x2+6x=6x+12.
$2x^2+6x$2x2+6x | $=$= | $6x+12$6x+12 |
$2x^2-12$2x2−12 | $=$= | $0$0 |
$2x^2$2x2 | $=$= | $12$12 |
$x^2$x2 | $=$= | $6$6 |
$\therefore$∴ $x$x | $=$= | $\sqrt{6}$√6, $x>0$x>0 |
As another example, consider a right-angled triangle with the two shorter sides given as $p^2-1$p2−1 and $2p$2p.
Can we find an expression for its perimeter and area?
By Pythagoras' theorem, we have that the hypotenuse $h$h is determined as:
$h^2$h2 | $=$= | $\left(p^2-1\right)^2+\left(2p\right)^2$(p2−1)2+(2p)2 |
$=$= | $p^4-2p^2+1+4p^2$p4−2p2+1+4p2 | |
$=$= | $p^4+2p^2+1$p4+2p2+1 | |
$=$= | $\left(p^2+1\right)^2$(p2+1)2 | |
$\therefore$∴ $h$h | $=$= | $p^2+1$p2+1 |
Thus the perimeter is given by the polynomial:
$P$P | $=$= | $\left(p^2+1\right)+\left(p^2-1\right)+2p$(p2+1)+(p2−1)+2p |
$=$= | $2p^2+2p$2p2+2p | |
$=$= | $2p\left(p+1\right)$2p(p+1) | |
The area is given by:
$A$A | $=$= | $\frac{1}{2}\left(2p\right)\left(p^2-1\right)$12(2p)(p2−1) |
$=$= | $\left(p\right)\left(p^2-1\right)$(p)(p2−1) | |
$=$= | $\left(p\right)\left(p+1\right)\left(p-1\right)$(p)(p+1)(p−1) |
Find the polynomial that represents the area of the triangle.
Jimmy and a few of his friends are jumping off their local wharf. If the length of the wharf is $\frac{6}{3y+2}$63y+2 metres and Jimmy jumps a distance of $\frac{5}{8y+8}$58y+8 metres, then how far does Jimmy travel from the start of his run up? Give your answer in factorised form.
Consider the cube shown.
Find the polynomial that represents the volume of the cube.
Find the volume of the cube if the value of $x$x is $6$6.