Q. 365 - Intermediate Algebra

Question

$3y^3-21y^2+30y$Factor Trinomials of the Form $a{x}^2+bx+c$ Using Trial and Error
In the following exercises, factor completely using trial and error.

$3y^3-21y^2+30y$

Step-by-Step Solution

Answer

By factoring the polynomial $3y^3-21y^2+30y$ completely by trial and error method, the factors are $3y(y-5)(y-2)$

Step 1. Given

The polynomial is $3y^3-21y^2+30y$

To factor the polynomial by trial and error method.

Step 2. Factor any GCF

Factor the GCF.

$3y^3-21y^2+30y=3y(y^2-7y+10)$

Step 3. Find all the pairs of factors of first term

The factors of the first term is $1,1$

Since there is only one term, we can put the in parentheses.
$(y )(y )$

Step 4. Find all the factor pairs of third term

Consider the sign.

Since the last term is $10$, the factors have same sign either positive or negative.

Since the sign of the second term is negative, both the factors have negative sign. 

The factors of $10$ are

• $-1,-10$
  
• $-2,-5$
  

Step 5. Test all combination of factors

Test all combination of factors until the correct product is found. 

Possible Factors	Product
$(y-1)(y-10)$	$y^2-11y+10$
$(y-2)(y-5)$	$y^2-7y+10$

The factors of polynomial is $y^2-7y+10=(y-2)(y-5)$

Step 6. Check the answer

Check by multiplying the factor. 

$3y(y-5)(y-2)=3y(y^2-7y+10)$

                        $=3y^3-21y^2+30y$