Q 104. - Intermediate Algebra

Question

Factor Trinomials of the Form $a{x}^2+bx+c$ Using Trial and Error.

$21m^2-29mn+10n^2$

Step-by-Step Solution

Answer

The solution is $(3m-2n)(7m-5n)$.

Step 1. Given information

Consider the trinomial.

$21m^2-29mn+10n^2$

Step 2. Write the trinomial in descending order and then find greatest common factor.

The trinomial $21m^2-29mn+10n^2$ is given in descending order.

There is no greatest common factor.

Step 3. Find the factors of the first term and the last term.

The first term of the given trinomial is $21m^2$.

So, the factors of $21m^2$ are as follows:

$\begin{array}{rcl}21m^2& =& m·21m\\ 21m^2& =& 3m·7m\end{array}$

The last term of the given trinomial is $10n^2$.

Since the middle term of the given polynomial is negative, the factors of the last term will both be negative.

The factors of $10n^2$ are as follows:

$\begin{array}{rcl}10n^2& =& (-n)·(-10n)\\ 10n^2& =& (-2n)·(-5n)\end{array}$

Step 4. Make a table for all the combination of factors of 21 m 2 − 29 m n + 10 n 2 .

If the trinomial has no common factors, none of the factors can contain the common factors.

This follows that the combination of the factors is not an option.

The table is shown below:

Possible factors	Product
$(m-n)(21m-10n)$	Not an option
$(m-10n)(21m-n)$	$21m^2-211mn+10n^2$
$(m-2n)(21m-5n)$	$21m^2-47mn+10n^2$
$(m-5n)(21m-2n)$	$21m^2-107mn+10n^2$
$(3m-n)(7m-10n)$	$21m^2-37mn+10n^2$
$(3m-10n)(7m-n)$	$21m^2-73mn+10n^2$
$(3m-2n)(7m-5n)$	$21m^2-29mn+10n^2$
$(3m-5n)(7m-2n)$	$21m^2-41mn+10n^2$

From the table, conclude that the combination $(3m-2n)(7m-5n)$ is correct.  

Step 5. Check by multiplying ( 3 m - 2 n ) ( 7 m - 5 n ) .

$\begin{array}{rcl}(3m-2n)(7m-5n)& =& 21m^2-15mn-14mn+10n^2\\ & =& 21m^2-29mn+10n^2\end{array}$

Hence, the factor is $(3m-2n)(7m-5n)$.