Question 407811




{{{28x^2-25x+3}}} as {{{(7x-3)(4x-1)}}}...let' multiply this

{{{(7x*4x+7x*(-1)-3*4x -3*(-1))}}}

{{{28x^2 -7x -12x +3}}}

{{{28x^2 -19x +3}}}...........Jacob was right


{{{28x^2-25x+3}}}....here Megan needed to replace {{{-25x}}} with sum or difference of two numbers in order to group terms (she needed four terms in this expression)

she took {{{-7x -12x}}} which makes {{{-19x}}}

she needed to replace {{{-25x}}} with a pair of numbers that add to {{{-25}}}

 first of all she needed to multiply the leading coefficient {{{28}}} and the last term {{{3}}},-she would have {{{84}}}

second of all she needed to list all of the possible factors of {{{84}}} and next

these factors pair up to multiply to 84 and see which of these pairs add to {{{-25}}}

1*84=84

2*42=84

3*28=84

4*21=84

6*14=84

7*12=84

(-1)*(-84)=84

(-2)*(-42)=84

(-3)*(-28)=84

(-4)*(-21)=84

(-6)*(-14)=84

(-7)*(-12)=84

We can see from the table that {{{-4}}} and {{{-21}}} add to {{{-25}}}; so, she needed them to use

let's try it:

 {{{28x^2-25x+3}}}

{{{28x^2 -4x -21x+3}}}...group terms

{{{(28x^2 -4x)+( -21x+3)}}}

{{{4x(7x -1)-3(7x-1)}}}...{{{(7x-1)}}} is a common term

{{{(4x-3)(7x-1)}}}.......this would be her correct solution