Lodshading

Factoring Expressions and Solving Equations1
Sample Solution 1. The command clear clears all variables. The command syms x declares x to be a symbolic variable. The command expr1 = (x-1)*(x-2)*(x-3)*(x-4)*(x-5) gives the label expr1 . to the expression The command expand is used to expand or multiply out an expression. Expanding expr1 yields The command factor is used to factor an expression. Here we factor the expression that results from expanding expr1. Thereby, we recover expr1, which is what one would expect. The relationship between solving and factoring is as follows: Let be any polynomial. is a solution of if and only if of . 2. M AT L AB is not able to factor
0 ¡ B edY¢XD G § ¡    5h¡ 2 B qpfpi2 G  2 ¡ 2 8 g52 ¡  8 gf2 ¡  6 g52 ¡  b ¡ ca`¡ G 6 gf2 ¡  0 ¡ § ¡   YXD § ¡   YXD C B  ¥ 7¨'7)@9%)7#&%(531)('&%¡ £ ¡  A  2 8 ¡ "   £ 6 ¡ " 4 2 0 ¡ " ¥ £ $ § " #! £ ¡   § C £ ¡   §  ¢©©¨¦¤¢  £ ¡   §  £ ¡   § ¥ £ ¡

Solving

gives

. symbolically; however, the solutions it

M AT L AB is able to solve gives are extremely long and complicated.
B riIPa2 G  2 ¡ 8 ge2 ¡  6 ge2 ¡  0 ¡

The command double(ans) numerically evaluates ans, in this case the symbolic solutions to . (Note. double(ans) does not mean 2 * ans; double is short for double precision.) The numerical solutions we obtain are
y u " u 4s B 77‚T@ƒ7‚€W)7xV@w77tVqP¡ v u 4  As ¥ £ R y " u " 4s B v u 4   s B G

One reason an exact, symbolic solution may not be as useful as an approximation is that when we measure things we usually use decimals or very simple fractions.
C 6 (q‰9qˆ(‡hg#2 § ¥ 2 ¡   ¡ G § ¥ 2 ¡  8 g52 ¡  6 9‡†2 ¡   ¡ G ¡ 8 g52 ¡  6 g#2 ¡  C

3. Solving is
0 ¡

gives

. The reason the answer is so nice is that .
B Q)1¡ G  E D

4. M AT L AB is unable to factor or to solve numerical solutions (to four decimal places)
C

, symbolically. M AT L AB gives the

None of the algorithms M AT L AB uses to obtain symbolic...