Question 119782

Now let's focus on the inner expression {{{w^2+4w-5}}}





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Looking at {{{w^2+4w-5}}} we can see that the first term is {{{w^2}}} and the last term is {{{-5}}} where the coefficients are 1 and -5 respectively.


Now multiply the first coefficient 1 and the last coefficient -5 to get -5. Now what two numbers multiply to -5 and add to the  middle coefficient 4? Let's list all of the factors of -5:




Factors of -5:

1,5


-1,-5 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -5

(1)*(-5)

(-1)*(5)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to 4? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 4


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-5</td><td>1+(-5)=-4</td></tr><tr><td align="center">-1</td><td align="center">5</td><td>-1+5=4</td></tr></table>



From this list we can see that -1 and 5 add up to 4 and multiply to -5



Now looking at the expression {{{w^2+4w-5}}}, replace {{{4w}}} with {{{-1w+5w}}} (notice {{{-1w+5w}}} adds up to {{{4w}}}. So it is equivalent to {{{4w}}})


{{{w^2+highlight(-1w+5w)+-5}}}



Now let's factor {{{w^2-1w+5w-5}}} by grouping:



{{{(w^2-1w)+(5w-5)}}} Group like terms



{{{w(w-1)+5(w-1)}}} Factor out the GCF of {{{w}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(w+5)(w-1)}}} Since we have a common term of {{{w-1}}}, we can combine like terms


So {{{w^2-1w+5w-5}}} factors to {{{(w+5)(w-1)}}}



So this also means that {{{w^2+4w-5}}} factors to {{{(w+5)(w-1)}}} (since {{{w^2+4w-5}}} is equivalent to {{{w^2-1w+5w-5}}})




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{{{-4w(w+5)(w-1)}}} Now reintroduce the outer term {{{-4w}}}


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Answer:


So {{{-4w(w^2+4w-5)}}} factors to {{{-4w(w+5)(w-1)}}}