How do you factor completely #-41x + 10 + 21x^2#?

The following is a lengthy, annoying method that doesn't require any guessing of factor pairs, as per the usual method of factoring:

Reorder the terms:

#p(x)=21x^2-41x+10#

Factor #21# from the first two terms.

#p(x)=21(x^2-41/21x)+10#

Complete the square inside the terms you previously factored the #21# from.

Since perfect squares come in the form #(x-a)^2=x^2-2ax+a^2#, we see that #2ax=41/21x#, we see that #a=41/42#. To make the term a perfect square, add the #a^2=(41/42)^2=1681/1764# term inside the parentheses.

Note that we can't add terms willy-nilly, so note that we will have to balance it on the outside.

#p(x)=21(x^2-41/21x+1681/1764)+10+?#

The #?# represents how we will balance the #a^2# we added. Note that the #1681/1764# we added is actually multiplied by the #21#, since it is in the parentheses, so its value is actually:

#(21xx1681)/1764=(3xx7xx41^2)/(2^2xx3^2xx7^2)=41^2/(2^2xx3xx7)=1681/84#

To balance the added #1684/84#, subtract its value as well.

#p(x)=21(x^2-41/21x+1681/1764)+10-1681/84#

Recall that #x^2-41/21x+1681/1764=(x-41/42)^2#.

Also note that #10-1681/84=840/84-1681/84=-841/84#.

#p(x)=21(x-41/42)^2-841/84#

We will factor this as a difference of squares. To do so, write both terms as squared terms.

#p(x)=(sqrt(21)(x-41/42))^2-(29/(2sqrt21))^2#

#p(x)=(sqrt21x-(41sqrt21)/42)^2-(29/(2sqrt21))^2#

Since #a^2-b^2=(a+b)(a-b)#:

#p(x)=(sqrt21x-(41sqrt21)/42+29/(2sqrt21))(sqrt21x-(41sqrt21)/42-29/(2sqrt21))#

Note that #-(41sqrt21)/42+29/(2sqrt21)=-(41sqrt21)/42+(29sqrt21)/42=-(12sqrt21)/42=-(2sqrt21)/7#.

Similarly, #-(41sqrt21)/42-29/(2sqrt21)=-(41sqrt21)/42-(29sqrt21)/42=-(70sqrt21)/42=-(5sqrt21)/3#

So:

#p(x)=(sqrt21x-(2sqrt21)/7)(sqrt21x-(5sqrt21)/3)#

Factoring #sqrt21# from both terms:

#p(x)=sqrt21(x-2/7)sqrt21(x-5/3)#

Since #sqrt21(sqrt21)=21#:

#p(x)=21(x-2/7)(x-5/3)#

Since #21=3xx7#:

#p(x)=7(x-2/7)*3(x-5/3)#

#p(x)=(7x-2)(3x-5)#

Re-arrange the terms into descending powers of x
#21x^2 -41x +10#

Combine factors of 21 and 10 in such away that their cross products add to 41.

#"7 2" rArr 6#
#"3 5" rArr 35 " "# #6+35 = 41#

The signs in the brackets will be the same because of the #color(red)(+21)#, they will both be negative, because of the #color(blue)(-41)#

The top row gives the one bracket, the bottom row gives the other bracket:

#(7x - 2)(3x -5)#

Use the new AC Method to factor trinomials (Socratic Search).
y = 21x^2 - 41x + 10 = 21(x + p)(x + q)
Converted trinomial: y' = x^2 - 41x + 210 = (x + p')(x + q').
Find p' and q' knowing they have same sign because ac > 0.
Factor pairs of (ac = 210) --> ...(5, 42)(-5, -42)(6, 35)(-6, -35). This sum is (-41 = b). Then, p' = -6 and q' = -35.
Back to original trinomial y --> p = (p')/a = -6/21 = - 2/7, and q = (q')/a = -35/21 = -5/3.
Factored form:
y = 21(x - 2/7)(x - 5/3) = (7x - 2)(3x - 5)