Question 326194
{{{2x^2+x}}}{{{""=""}}}{{{42+9x}}}
<pre><b>

The other tutor's solution is correct; however there is one step
he did not do, which would have made the factoring easier.

Get a zero on the right by adding {{{red(-42-9x)}}} to both sides

{{{2x^2+x}}}{{{red(-42-9x)}}}{{{""=""}}}{{{42+9x}}}{{{red(-42-9x)}}}

Combine like terms and simplify:

{{{2x^2}}}{{{""-""}}}{{{8x}}}{{{""-""}}}{{{42}}}{{{""=""}}}{{{0}}}

Since all the numbers are even we can divide every term by {{{red(2)}}}

{{{2x^2/red(2)}}}{{{""-""}}}{{{8x/red(2)}}}{{{""-""}}}{{{42/red(2)}}}{{{""=""}}}{{{0/red(2)}}}

[That is the step which the other tutor did not do, which makes it easier
to factor.]

Simplify:

{{{x^2}}}{{{""-""}}}{{{4*x}}}{{{""-""}}}{{{red(21)}}}{{{""=""}}}{{{0}}}

We must factor that.

I did some coloring so I can refer to the parts:

List all the ways to factor the red number 21 using two factors,
(list the larger factor first):

{{{21}}}{{{x}}}{{{1}}}
 {{{7}}}{{{x}}}{{{3}}}

{{{x^2}}}{{{""-""}}}{{{4*x}}}{{{red(""-"")}}}{{{21}}}{{{""=""}}}{{{0}}}

Notice that the red sign (the one just before the 21 is {{{red(""-"")}}}

Put that sign between those numbers above instead of the multiplication 
sign, and give the answer:

{{{21}}}{{{""-""}}}{{{1}}}{{{""=""}}}{{{20}}}
 {{{7}}}{{{""-""}}}{{{3}}}{{{""=""}}}{{{red(4)}}}

Notice that the {{{red(4)}}} is the same as the middle coefficient
ignoring the sign: 

{{{x^2}}}{{{""-""}}}{{{red(4)x}}}{{{""-""}}}{{{21}}}{{{""=""}}}{{{0}}}

Therefore write this:

{{{"(x"}}}{{{""}}}{{{""}}}{{{"7)"}}}{{{"(x"}}}{{{""}}}{{{""}}}{{{"3)"}}}{{{""=""}}}{{{0}}}

Finally we put the signs in.  7 is the larger factor, and the red
sign (the sign before the {{{4*x}}} is {{{red(""-"")}}} in this:

{{{x^2}}}{{{red(""-"")}}}{{{4*x}}}{{{""-""}}}{{{21}}}{{{""=""}}}{{{0}}}

So we place that sign before the 7

{{{"(x"}}}{{{red(""-"")}}}{{{"7)"}}}{{{"(x"}}}{{{""}}}{{{""}}}{{{"3)"}}}{{{""=""}}}{{{0}}}

Next we place whichever sign before the 3 which will cause the product
to come out to be {{{-21}}}, the last term of {{{x^2}}}{{{""-""}}}{{{4*x}}}{{{""-""}}}{{{21}}}{{{""=""}}}{{{0}}}

That sign would have to be {{{""+""}}}, so we have
now completely factored the left side:

{{{"(x"}}}{{{red(""-"")}}}{{{"7)"}}}{{{"(x"}}}{{{"+"}}}{{{"3)"}}}{{{""=""}}}{{{0}}}

Next we use the zero-factor principle.

Each of the factors {{{"(x"}}}{{{""-""}}}{{{"7)"}}} and {{{"(x"}}}{{{""+"")}}}{{{"3)"}}} represents some number.  
Since their product must equal to 0,
it is obvious that one of them is equal to 0.  But which one?
We cannot tell, so it would be correct if {{{"(x"}}}{{{""-""}}}{{{"7)"}}} were
0 and {{{"(x"}}}{{{""+""}}}{{{"3)"}}} something else. OR, it would also
be correct if {{{"(x"}}}{{{""+""}}}{{{"3)"}}} were 0 and {{{"(x"}}}{{{""-""}}}{{{"7)"}}} were something else. 

That's why there are two solutions to a quadratic equation. So we
have to consider both possibilities:

{{{x}}}{{{""-""}}}{{{7}}}{{{""=""}}}{{{0}}} OR {{{x}}}{{{""+""}}}{{{3}}}{{{""=""}}}{{{0}}}

Solvving those we have two possibilities for solutions:

{{{x}}}{{{""=""}}}{{{7}}} OR {{{x}}}{{{""=""}}}{{{-3}}}

Edwin</pre>