Page:Elementary algebra (1896).djvu/25

Rh given, and by combining the terms the numerical value of the whole expression is obtained.

16. We have already, in Art. S 8, called attention to the importance of carefully distinguishing between coefficient and index; confusion between these is such a fruitful source of error with beginners that it may not be unnecessary once more to dwell on the distinction.

{{fine block|{{sc|Ex}}. 1. When $$c = 5$$, find the value of $$c^4 - 4c + 22c^3 - 3c^2$$.

{{float left|Here}} {{block center|$$\begin{align} c^4 & = 5^4 = 5 \times 5 \times 5 \times 5 \times 5 = 625 \\ 4c & = 4 \times 5 = 20 \\ 2c^3 & = 2 \times 5^3 = 2 \times 5 \times 5 \times 5 = 250 \\ 3c^2 & = 3 \times 5^2 - 3 \times 5 \times 5 = 75 \end{align}$$}}

Hence the value of the expression $$= 625 - 20 + 250 - 75 = 780$$.

{{sc|Ex}}. 2. When $$p = 9$$, $$r = 6$$, $$k = 4$$, find the value of

{{c|$$\frac{1}{3}\sqrt[3]{\left(\frac{pr}{k^2}\right)} + \sqrt{(3k + k^3 + 5)} - \frac{2r^2}{9k}$$.}}

{{block center|$$\begin{align} \frac{1}{3}\sqrt[3]{\left(\frac{pr}{k^2}\right)} + \sqrt{(3k + k^3 + 5)} - \frac{2r^2}{9k} & = \frac{1}{3}\sqrt[3]{\left(\frac{54}{16}\right)} + \sqrt{(12+64+5)} - \frac{2\times 36}{9\times 4} \\ & = \frac{1}{3}\sqrt[3]{\frac{27}{8}} + \sqrt{81} - 2 \\ & = \frac{1}{3}\times \frac{3}{2} + 9 - 2 = 7\frac{1}{2} \end{align}$$}}

{{larger|17}}. By Art. 13 any term which contains a zero factor is itself zero, and may be called a zero term.

{{fine block|{{sc|Ex.}} 1. If $$a = 2$$, $$b = 0$$, $$x = 3$$, $$y = 1$$, find the value of

{{c|$$4a^3 - ab^3 + 2xy^2 + 3abx$$}}

{{float left|The expression}}{{block center|$$\begin{align}&=(4\times 2^3) - 0 + (2\times 3\times 1) + 0\\ &=32-0+6+0=38\end{align}$$ }}

{{sc|Note.}} The two zero terms do not affect the result.

{{sc|Ex.}} 2. Find the value of $$\frac{3}{5}x^2-a^2y+7abx-\frac{5}{2}y^3$$, when

{{c|$$a = 5$$, $$b = 0$$, $$x = 7$$, $$y = 1$$.}}

{{block center|$$\begin{align}\frac{3}{5}x^2-a^2y+7abx-\frac{5}{2}y^3 & =\frac{3}{5}\times 7^2 - 5^2 \times 1 + 0 - \frac{5}{2} \times 1^3 \\ & =29\frac{3}{5} = 25 - 2\frac{1}{2}=1\frac{9}{10}\end{align} $$}}

{{sc|Note.}} The zero term does not affect the result.}}

{{larger|18}}. In working examples the student should pay attention to the following hints:

1. Too much importance cannot be attached to neatness of style and arrangement. The beginner should remember that neatness is in itself conducive to accuracy. {{nop}}