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)