Question 87196
In order to factor {{{12*w^2+19*w+4}}}, first multiply 12 and 4 to get 48 and we need to ask ourselves: What two numbers multiply to 48 and add to 19? Lets find out by listing all of the possible factors of 48


Factors:

1,2,3,4,6,8,12,16,24,48,

-1,-2,-3,-4,-6,-8,-12,-16,-24,-48, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 48.

1*48=48

2*24=48

3*16=48

4*12=48

6*8=48

(-1)*(-48)=48

(-2)*(-24)=48

(-3)*(-16)=48

(-4)*(-12)=48

(-6)*(-8)=48

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

 Now which of these pairs add to 19? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 19

<TABLE><TR><TD>First Number</TD><TD>|</TD><TD>Second Number</TD><TD>|</TD><TD>Sum</TD></TR><TR><TD>1</TD><TD>|</TD><TD>48</TD>|</TD><TD>|</TD><TD>1+48=49</TD></TR><TR><TD>2</TD><TD>|</TD><TD>24</TD>|</TD><TD>|</TD><TD>2+24=26</TD></TR><TR><TD>3</TD><TD>|</TD><TD>16</TD>|</TD><TD>|</TD><TD>3+16=19</TD></TR><TR><TD>4</TD><TD>|</TD><TD>12</TD>|</TD><TD>|</TD><TD>4+12=16</TD></TR><TR><TD>6</TD><TD>|</TD><TD>8</TD>|</TD><TD>|</TD><TD>6+8=14</TD></TR><TR><TD>-1</TD><TD>|</TD><TD>-48</TD>|</TD><TD>|</TD><TD>-1+(-48)=-49</TD></TR><TR><TD>-2</TD><TD>|</TD><TD>-24</TD>|</TD><TD>|</TD><TD>-2+(-24)=-26</TD></TR><TR><TD>-3</TD><TD>|</TD><TD>-16</TD>|</TD><TD>|</TD><TD>-3+(-16)=-19</TD></TR><TR><TD>-4</TD><TD>|</TD><TD>-12</TD>|</TD><TD>|</TD><TD>-4+(-12)=-16</TD></TR><TR><TD>-6</TD><TD>|</TD><TD>-8</TD>|</TD><TD>|</TD><TD>-6+(-8)=-14</TD></TR><TABLE>We can see from the table that 3 and 16 add to 19. So the two numbers that multiply to 48 and add to 19 are: 3 and 16

 So the original quadratic


{{{12*w^2+19*w+4}}}


breaks down to this (just replace {{{19*w}}} with the two numbers that multiply to 48 and add to 19, which are: 3 and 16)


 {{{12*w^2+3w+16w+4}}}

Group the first two terms together and the last two terms together like this:

{{{(12*w^2+3w)+(16w+4)}}}

Factor a 3 out of the first group and factor a 4 out of the second group.


{{{3w(4w+1)+4(4w+1)}}}


Now since we have a common term {{{4w+1}}} we can combine the two terms. Notice if we let {{{y=4w+1}}} we would get {{{3wy+4y=(3w+4)y}}}. Since we have that common term {{{y}}}, we are able to combine {{{3w}}} and {{{4}}}


{{{(3w+4)(4w+1)}}} Combine like terms. 

Answer:

So the quadratic {{{12*w^2+19*w+4}}} factors to


{{{(3w+4)(4w+1)}}}



Notice how {{{(3w+4)(4w+1)}}} foils back to our original problem {{{12*w^2+19*w+4}}}. This verifies our answer.





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{{{4z^3-18z^2-10z }}}


{{{z(4z^2-18z-10) }}} Factor out a "z"


Now lets factor the quadratic inside the parenthesis


In order to factor {{{4*z^2-18*z-10}}}, first multiply 4 and -10 to get -40 and we need to ask ourselves: What two numbers multiply to -40 and add to -18? Lets find out by listing all of the possible factors of -40


Factors:

1,2,4,5,8,10,20,40,

-1,-2,-4,-5,-8,-10,-20,-40, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to -40.

(-1)*(40)=-40

(-2)*(20)=-40

(-4)*(10)=-40

(-5)*(8)=-40

 Now which of these pairs add to -18? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -18

<TABLE><TR><TD>First Number</TD><TD>|</TD><TD>Second Number</TD><TD>|</TD><TD>Sum</TD></TR><TR><TD>1</TD><TD>|</TD><TD>-40</TD>|</TD><TD>|</TD><TD>1+(-40)=-39</TD></TR><TR><TD>2</TD><TD>|</TD><TD>-20</TD>|</TD><TD>|</TD><TD>2+(-20)=-18</TD></TR><TR><TD>4</TD><TD>|</TD><TD>-10</TD>|</TD><TD>|</TD><TD>4+(-10)=-6</TD></TR><TR><TD>5</TD><TD>|</TD><TD>-8</TD>|</TD><TD>|</TD><TD>5+(-8)=-3</TD></TR><TR><TD>-1</TD><TD>|</TD><TD>40</TD>|</TD><TD>|</TD><TD>(-1)+40=39</TD></TR><TR><TD>-2</TD><TD>|</TD><TD>20</TD>|</TD><TD>|</TD><TD>(-2)+20=18</TD></TR><TR><TD>-4</TD><TD>|</TD><TD>10</TD>|</TD><TD>|</TD><TD>(-4)+10=6</TD></TR><TR><TD>-5</TD><TD>|</TD><TD>8</TD>|</TD><TD>|</TD><TD>(-5)+8=3</TD></TR><TABLE>We can see from the table that 2 and -20 add to -18. So the two numbers that multiply to -40 and add to -18 are: 2 and -20

 So the original quadratic


{{{4*z^2-18*z-10}}}


breaks down to this (just replace {{{-18*z}}} with the two numbers that multiply to -40 and add to -18, which are: 2 and -20)


 {{{4*z^2+2z-20z-10}}}

Group the first two terms together and the last two terms together like this:

{{{(4*z^2+2z)+(-20z-10)}}}

Factor a 2 out of the first group and factor a -10 out of the second group.


{{{2z(2z+1)-10(2z+1)}}}


Now since we have a common term {{{2z+1}}} we can combine the two terms. Notice if we let {{{y=2z+1}}} we would get {{{2zy-10y=(2z-10)y}}}. Since we have that common term {{{y}}}, we are able to combine {{{2z}}} and {{{-10}}}


{{{(2z-10)(2z+1)}}} Combine like terms. 

Answer:

So the quadratic {{{4*z^2-18*z-10}}} factors to


{{{(2z-10)(2z+1)}}}



Notice how {{{(2z-10)(2z+1)}}} foils back to our original problem {{{4*z^2-18*z-10}}}. This verifies our answer.




So now reintroduce the z term we factored out back in:


{{{z(2z-10)(2z+1)}}}


So the expression 


{{{4z^3-18z^2-10z }}}


factors to this


{{{z(2z-10)(2z+1)}}}


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