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

Jun 21, 2016

$\left(7 x - 2\right) \left(3 x - 5\right)$

#### Explanation:

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:

$p \left(x\right) = 21 {x}^{2} - 41 x + 10$

Factor $21$ from the first two terms.

$p \left(x\right) = 21 \left({x}^{2} - \frac{41}{21} x\right) + 10$

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

Since perfect squares come in the form ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$, we see that $2 a x = \frac{41}{21} x$, we see that $a = \frac{41}{42}$. To make the term a perfect square, add the ${a}^{2} = {\left(\frac{41}{42}\right)}^{2} = \frac{1681}{1764}$ 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 $\frac{1681}{1764}$ we added is actually multiplied by the $21$, since it is in the parentheses, so its value is actually:

$\frac{21 \times 1681}{1764} = \frac{3 \times 7 \times {41}^{2}}{{2}^{2} \times {3}^{2} \times {7}^{2}} = {41}^{2} / \left({2}^{2} \times 3 \times 7\right) = \frac{1681}{84}$

To balance the added $\frac{1684}{84}$, subtract its value as well.

$p \left(x\right) = 21 \left({x}^{2} - \frac{41}{21} x + \frac{1681}{1764}\right) + 10 - \frac{1681}{84}$

Recall that ${x}^{2} - \frac{41}{21} x + \frac{1681}{1764} = {\left(x - \frac{41}{42}\right)}^{2}$.

Also note that $10 - \frac{1681}{84} = \frac{840}{84} - \frac{1681}{84} = - \frac{841}{84}$.

$p \left(x\right) = 21 {\left(x - \frac{41}{42}\right)}^{2} - \frac{841}{84}$

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

$p \left(x\right) = {\left(\sqrt{21} \left(x - \frac{41}{42}\right)\right)}^{2} - {\left(\frac{29}{2 \sqrt{21}}\right)}^{2}$

$p \left(x\right) = {\left(\sqrt{21} x - \frac{41 \sqrt{21}}{42}\right)}^{2} - {\left(\frac{29}{2 \sqrt{21}}\right)}^{2}$

Since ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$:

$p \left(x\right) = \left(\sqrt{21} x - \frac{41 \sqrt{21}}{42} + \frac{29}{2 \sqrt{21}}\right) \left(\sqrt{21} x - \frac{41 \sqrt{21}}{42} - \frac{29}{2 \sqrt{21}}\right)$

Note that $- \frac{41 \sqrt{21}}{42} + \frac{29}{2 \sqrt{21}} = - \frac{41 \sqrt{21}}{42} + \frac{29 \sqrt{21}}{42} = - \frac{12 \sqrt{21}}{42} = - \frac{2 \sqrt{21}}{7}$.

Similarly, $- \frac{41 \sqrt{21}}{42} - \frac{29}{2 \sqrt{21}} = - \frac{41 \sqrt{21}}{42} - \frac{29 \sqrt{21}}{42} = - \frac{70 \sqrt{21}}{42} = - \frac{5 \sqrt{21}}{3}$

So:

$p \left(x\right) = \left(\sqrt{21} x - \frac{2 \sqrt{21}}{7}\right) \left(\sqrt{21} x - \frac{5 \sqrt{21}}{3}\right)$

Factoring $\sqrt{21}$ from both terms:

$p \left(x\right) = \sqrt{21} \left(x - \frac{2}{7}\right) \sqrt{21} \left(x - \frac{5}{3}\right)$

Since $\sqrt{21} \left(\sqrt{21}\right) = 21$:

$p \left(x\right) = 21 \left(x - \frac{2}{7}\right) \left(x - \frac{5}{3}\right)$

Since $21 = 3 \times 7$:

$p \left(x\right) = 7 \left(x - \frac{2}{7}\right) \cdot 3 \left(x - \frac{5}{3}\right)$

$p \left(x\right) = \left(7 x - 2\right) \left(3 x - 5\right)$

Jun 21, 2016

$\left(7 x - 2\right) \left(3 x - 5\right)$

#### Explanation:

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

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

$\text{7 2} \Rightarrow 6$
$\text{3 5" rArr 35 " }$ $6 + 35 = 41$

The signs in the brackets will be the same because of the $\textcolor{red}{+ 21}$, they will both be negative, because of the $\textcolor{b l u e}{- 41}$

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

$\left(7 x - 2\right) \left(3 x - 5\right)$

Jun 22, 2016

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

#### Explanation:

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)