(A) 9x²-13x-10
Multiply the 9 by the 10 ignoring signs. Get 90
Write down all the ways to have two positive integers
which have product 90, starting with 90*1
90*1
45*2
30*3
18*5
15*6
10*9
Since the last sign in 9x²-13x-10 is -, SUBTRACT them,
and place the DIFFERENCE out beside that:
90*1 90-1=89
45*2 45-2=44
30*3 30-3=27
18*5 18-5=13
15*6 15-6= 9
10*9 10-9= 1
Now, again ignoring signs, we find in that list of
sums the coefficient of the middle term in 9x²-13x-10
So we replace the number 13 by 18-5
9x²-13x-10
9x²-(18-5)x-10
Then we distribute to remove the parentheses:
9x²-18x+5x-10
Factor the first two terms 9x²-18 by taking out the
greatest common factor, getting 9x(x-2)
Factor the last two terms +5x-10 by taking out the
greatest common factor, getting +5(x-2)
So we have
9x(x-2)+5(x-2)
Notice that there is a common factor, (x-2)
9x(x-2)+5(x-2)
which we can factor out leaving the 9x and the +5 to put
in parentheses:
(x-2)(9x+5)
---------------------------
(B) 12x²-32x+21
Multiply the 12 by the 21 ignoring signs. Get 252
Write down all the ways to have two positive integers
which have product 252, starting with 252*1
252*1
126*2
84*3
63*4
42*6
36*7
28*9
21*12
18*14
Since the last sign in 12x²-32x+21 is +, ADD them,
and place the SUM out beside that:
252*1 252+1=253
126*2 126+2=128
84*3 84+3-87
63*4 63+4=67
42*6 42+6=48
36*7 36+7=43
28*9 28+9=37
21*12 21+12=33
18*14 18+14=32
Now, again ignoring signs, we find in that list of
sums the coefficient of the middle term in 12x²-32x+21
So we replace the number 32 by 18+14
12x²-32x+21
12x²-(18+14)x+21
Then we distribute to remove the parentheses:
12x²-18x-14x+21
Factor the first two terms 12x²-18x by taking out the
greatest common factor, getting 6x(2x-3)
Factor the last two terms -14x+21 by taking out the
greatest common factor, getting -7(2x-3)
So we have
6x(2x-3)-7(2x-3)
Notice that there is a common factor, (2x-3)
6x(2x-3)-7(2x-3)
which we can factor out leaving the 6x and the -7 to put
in parentheses:
(2x-3)(6x-7)
Edwin
