# Khan/khan-exercises

1 parent 2a4d0c8 commit 2f417d3a58f9c26eb421ec9457e474cba141114b mathiash committed Apr 13, 2012
 @@ -52,12 +52,6 @@
a factored expression, like (x+1)(x+3)
-
 @@ -86,15 +86,19 @@

or

parseFormat("(#{" + A + "}x^2+#{" + a + "}x)+(#{" + b + "}x+#{" + C + "})", [BLUE, GREEN, GREEN, ORANGE])

The next step is to factor both terms of the above expression:

-

- format(HINT1, {del1factors:true, evalBasicNumOps:true}) -

-

- Redistribute the common term to get the answer: -

-

- format(SOLUTION, simplifyOptions.basic, false, "large") -

+
+

+ format(HINT1, {del1factors:true, evalBasicNumOps:true}) +

+
+
+

+ Redistribute the common term to get the answer: +

+

+ format(SOLUTION, simplifyOptions.basic, false, "large") +

+
 @@ -178,10 +178,10 @@ var hintExpr = getFractionFromOccFactors(factors, newOccFactors, argsOccFactors); if (KhanUtil.exprIdentical(newExpr, solExpr)) { - steps.add("

There are no factors that can be simplified in this expression, so the answer is: " + KhanUtil.format(solExpr) + ""); + steps.add("

There are no factors that can be simplified in this expression, so the answer is: " + KhanUtil.format(solExpr) + ""); } else { steps.add("

Applying the approach described above gives in this case:

" + KhanUtil.format(hintExpr) + "

We obtain the following expression:

" + KhanUtil.format(solExpr) + "