SOLUTION: {{{a^2+13a+36=0}}}

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Question 77047This question is from textbook Glencoe Algebra 1
: This question is from textbook Glencoe Algebra 1

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!

Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)

In order to factor , first we need to ask ourselves: What two numbers multiply to 36 and add to 13? Lets find out by listing all of the possible factors of 36

Factors:

1,2,3,4,6,9,12,18,36,

-1,-2,-3,-4,-6,-9,-12,-18,-36,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 36.

1*36=36

2*18=36

3*12=36

4*9=36

6*6=36

(-1)*(-36)=36

(-2)*(-18)=36

(-3)*(-12)=36

(-4)*(-9)=36

(-6)*(-6)=36

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 13? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 13

||||||||||
First Number | Second Number | Sum
1 | 36 | 1+36=37
2 | 18 | 2+18=20
3 | 12 | 3+12=15
4 | 9 | 4+9=13
6 | 6 | 6+6=12
-1 | -36 | -1+(-36)=-37
-2 | -18 | -2+(-18)=-20
-3 | -12 | -3+(-12)=-15
-4 | -9 | -4+(-9)=-13
-6 | -6 | -6+(-6)=-12
We can see from the table that 4 and 9 add to 13. So the two numbers that multiply to 36 and add to 13 are: 4 and 9 Now we substitute these numbers into a and b of the general equation of a product of linear factors which is: substitute a=4 and b=9 So the equation becomes: (x+4)(x+9) Notice that if we foil (x+4)(x+9) we get the quadratic again

Just replace the x's with a's to get:

Set each term to zero

So our answer is
or