Question 249231
how do you factor 

{{{x^2-9x+20}}}
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We try to factor it.

There is a 1 in front of the {{{x^2}}} and a POSITIVE
number on the right.

{{{1x^2-9x+20}}}

Multiply the 1 by the 20, get 20.  Write down all the ways
you can multiply two positive integers together to get 20.
They are

1x20
2x10
4x5

Now since the last sign before the 20 is +, out to the
right of those we ADD those pairs of positive integers:

1x20   20+1=21
2x10   10+2=12
4x5     4+5=9

Ignoring signs we see that 9 is the coefficient of the
middle term, and that corresponds to factors 4 and 5,
So we must put signs, either + or -, in front of 4x and 5x 
so they will combine to give the middle term, -9x.  We 
can do that by assigning a negative sign to 4x and a 
negative sign to 5x.  So the middle term -9x can be
written as -4x-5x
 
So

{{{x^2-9x+20}}}

becomes

{{{x^2-4x-5x+20}}}

Factor only the first two terms:

{{{x(x-4)-5x+20}}}

Factor the last two terms:

{{{x(x-4)-5(x-4)}}}

Now notice there is a common binomial factor of
{{{(x-4)}}}, so we can take it out:

{{{(x-4)(x-5)}}}



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and How Do You Solve For The Variable in 
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2x^2+x-2=0

We first try to factor the left side.

There is a 1 in front of the {{{x}}} and a NEGATIVE
number on the right.

{{{2x^2+1x-2=0}}}

Multiply the 2 by the 2, get 4.  Write down all the ways
you can multiply two positive integers together to get 4.
They are

1x4
2x2

Now since the last sign before the second 2 is -, out to the
right of those we SUBTRACT those pairs of positive integers:

1x4   4-1=3
2x2   2-2=0

We see that the coefficient of the middle term, 1, ignoring
signs, is neither 3 nor 0.  That means the trinomial will not 
factor.  

So we must use the quadratic formula:

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

with {{{a=2}}}, {{{b=1}}},  {{{c=-2}}}

{{{x = (-(1) +- sqrt((1)^2-4*(2)*(-2)))/(2*(2)) }}}

{{{x = (-1 +- sqrt(1+16))/4 }}}

{{{x = (-1 +- sqrt(17))/4 }}}

Edwin</pre>