Question 620914
This is how I do it.
To factor the quadratic trinomial {{{10x^2-7x-3}}} ,
I first multiply together the absolute values of the leading coefficient (10) and the independent term (-3) to get {{{10*3=30}}}.
I look for, and list all pairs of factors that multiply to yield that product, and find 4 such pairs:
{{{1*30=30}}}
{{{2*15=30}}}
{{{3*10=30}}}
{{{5*6=30}}}.
One of those pairs will be the absolute values of the coefficients of first degree terms obtained when multiplying the final factorization.
Because the independent term (-3) is negative, I know that those coefficients of first degree terms have opposite signs.
I also know that they add up to the coefficient of the first degree term in the original polynomial (-7).
From the four pairs of factors found above, the pair of factors, with signs, that add up to -7 is
{{{-10+3=-7}}}.
The expanded product of the factorization will contain {{{-10x}}} and {{{+3x}}} in addition to the leading and independent terms {{{10x^2}}} and {{{-3}}}.
I then organize the expanded product of the factorization in a 2 by 2 square, with the newly found terms at opposite corners:
{{{drawing(150,150,-11,11,-11,11,
rectangle(10,10,-10,-10),
line(-10,0,10,0), line(0,-10,0,10),
locate(-7,7,10x^2), locate(2.5,6,-10x),
locate(-6,-4,+3x), locate(3.5,-4,-3)
)}}}
Next, I look for common factors for each row and column and write them on the same row/column, outside the square.
I find {{{10x}}} as a common factor for {{{10x^2}}} and {{{-10x}}}, so I write {{{10x}}} to the left of {{{10x^2}}}.
I find {{{3}}} as a common factor for {{{3x}}} and {{{-3}}}, so I write {{{3}}} to the left of {{{3x}}}.
I write {{{x}}} above {{{10x^2}}} and {{{-10x}}}, because it is their common factor.
I write {{{-1}}} above {{{-10x}}} and {{{-3}}}, because it is their common factor.
{{{drawing(175,175,-15,11,-11,15,
rectangle(10,10,-10,-10),
line(-10,0,10,0), line(0,-10,0,10),
locate(-7,7,10x^2), locate(2.5,6,-10x),
locate(-6,-4,+3x), locate(3.5,-4,-3),
locate(-14,6,10x), locate(-13,-4,+3),
locate(-4.5,14,x), locate(3.5,14,-1)
)}}}
Now I make a binomial of the terms written above the square {{{x-1}}} and another binomial with the terms written to the left of the square {{{10x+3}}}.
I multiply those binomials to verify that the terms inside the square are generated.
I also verify that collecting terms in that product produces the original trinomial.
{{{(10x+3)(x-1)=10x^2-10x+3x-3=10x^2-7x-3}}}