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:
Example 1:
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 |
Example 2:
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) |
Worked Examples
Question 1
Find the polynomial that represents the area of the triangle.
Question 2
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 factored form.
Question 3
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.