Question 196046
I'm assuming that you want to factor {{{x^2 - 12x + 35}}}





Looking at the expression {{{x^2-12x+35}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-12}}}, and the last term is {{{35}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{35}}} to get {{{(1)(35)=35}}}.



Now the question is: what two whole numbers multiply to {{{35}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-12}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{35}}} (the previous product).



Factors of {{{35}}}:

1,5,7,35

-1,-5,-7,-35



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{35}}}.

1*35
5*7
(-1)*(-35)
(-5)*(-7)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-12}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>35</font></td><td  align="center"><font color=black>1+35=36</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>7</font></td><td  align="center"><font color=black>5+7=12</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-35</font></td><td  align="center"><font color=black>-1+(-35)=-36</font></td></tr><tr><td  align="center"><font color=red>-5</font></td><td  align="center"><font color=red>-7</font></td><td  align="center"><font color=red>-5+(-7)=-12</font></td></tr></table>



From the table, we can see that the two numbers {{{-5}}} and {{{-7}}} add to {{{-12}}} (the middle coefficient).



So the two numbers {{{-5}}} and {{{-7}}} both multiply to {{{35}}} <font size=4><b>and</b></font> add to {{{-12}}}



Now replace the middle term {{{-12x}}} with {{{-5x-7x}}}. Remember, {{{-5}}} and {{{-7}}} add to {{{-12}}}. So this shows us that {{{-5x-7x=-12x}}}.



{{{x^2+highlight(-5x-7x)+35}}} Replace the second term {{{-12x}}} with {{{-5x-7x}}}.



{{{(x^2-5x)+(-7x+35)}}} Group the terms into two pairs.



{{{x(x-5)+(-7x+35)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x-5)-7(x-5)}}} Factor out {{{7}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(x-7)(x-5)}}} Combine like terms. Or factor out the common term {{{x-5}}}


---------------------------------------------



Answer:



So {{{x^2-12x+35}}} factors to {{{(x-7)(x-5)}}}.



Note: you can check the answer by FOILing {{{(x-7)(x-5)}}} to get {{{x^2-12x+35}}} or by graphing the original expression and the answer (the two graphs should be identical).




<hr>




# 2





Looking at the expression {{{2x^2+11x+5}}}, we can see that the first coefficient is {{{2}}}, the second coefficient is {{{11}}}, and the last term is {{{5}}}.



Now multiply the first coefficient {{{2}}} by the last term {{{5}}} to get {{{(2)(5)=10}}}.



Now the question is: what two whole numbers multiply to {{{10}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{11}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{10}}} (the previous product).



Factors of {{{10}}}:

1,2,5,10

-1,-2,-5,-10



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{10}}}.

1*10
2*5
(-1)*(-10)
(-2)*(-5)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{11}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=red>1</font></td><td  align="center"><font color=red>10</font></td><td  align="center"><font color=red>1+10=11</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>2+5=7</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>-1+(-10)=-11</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-2+(-5)=-7</font></td></tr></table>



From the table, we can see that the two numbers {{{1}}} and {{{10}}} add to {{{11}}} (the middle coefficient).



So the two numbers {{{1}}} and {{{10}}} both multiply to {{{10}}} <font size=4><b>and</b></font> add to {{{11}}}



Now replace the middle term {{{11x}}} with {{{x+10x}}}. Remember, {{{1}}} and {{{10}}} add to {{{11}}}. So this shows us that {{{x+10x=11x}}}.



{{{2x^2+highlight(x+10x)+5}}} Replace the second term {{{11x}}} with {{{x+10x}}}.



{{{(2x^2+x)+(10x+5)}}} Group the terms into two pairs.



{{{x(2x+1)+(10x+5)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(2x+1)+5(2x+1)}}} Factor out {{{5}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(x+5)(2x+1)}}} Combine like terms. Or factor out the common term {{{2x+1}}}


---------------------------------------------



Answer:



So {{{2x^2+11x+5}}} factors to {{{(x+5)(2x+1)}}}.



Note: you can check the answer by FOILing {{{(x+5)(2x+1)}}} to get {{{2x^2+11x+5}}} or by graphing the original expression and the answer (the two graphs should be identical).





<hr>



# 3


I'm assuming you want to factor {{{x^4y - 16y}}}



{{{x^4y - 16y}}} Start with the given expression



{{{y(x^4 - 16)}}} Factor out the GCF



{{{y(x^2 - 4)(x^2 +4)}}} Factor {{{x^4 - 16}}} using the difference of squares formula



{{{y(x-2)(x+2)(x^2 +4)}}} Factor {{{x^2 - 4}}} using the difference of squares formula



So {{{x^4y - 16y}}} completely factors to {{{y(x-2)(x+2)(x^2 +4)}}}