Question

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

$10y^2-53y-11$

Short answer

The solution is $(2y-11)(5y+1)$.

Step-by-step solution

Step 1. Given information

Consider the trinomial.

$10y^2-53y-11$

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

The trinomial $10y^2-53y-11$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 $10y^2$.

So, the factors of $10y^2$are as follows:

$$\begin{gathered} 10y^2=y·10y \\ 10y^2=2y·5y \end{gathered}$$

The last term of the given trinomial is $-11$.

Since the last term of the given polynomial is negative, the factors of the last term will have opposite signs.

The factors of $-11$are as follows:

$\begin{array}{rcl}-11& =& 1·(-11)\\ -11& =& (-1)·11\end{array}$

Step 4. Make a table for all the combination of factors of 10y2−53y−11. 

Possible factors	Product
$(y+1)(10y-11)$	$10y^2-y-11$
$(y-11)(10y+1)$	$10y^2-109y-11$
$(y-1)(10y+11)$	$10y^2+y-11$
$(y+11)(10y-1)$	$10y^2+109y-11$
$(2y+1)(5y-11)$	$10y^2-17y-11$
$(2y-11)(5y+1)$	$10y^2-53y-11$
$(2y-1)(5y+11)$	$10y^2+17y-11$
$(2y+11)(5y-1)$	$10y^2+53y-11$

From the table, conclude that the combination localid="1644583284895" $(2y-11)(5y+1)$is correct.

Step 5. Check by multiplying (2y-11)(5y+1).

$\begin{array}{rcl}(2y-11)(5y+1)& =& 10y^2+2y-55y-11\\ & =& 2y^2-53y-11\end{array}$

Hence, the factor is $(2y-11)(5y+1)$.