Factoring Polynomials

Factoring Polynomials

Problem No. 98

 

3h²t +6ht +3t

This problem will be solved by the method of factoring out the Greatest Common Factor (GCF)

Step 1: Simplify the polynomial by factoring the GCF first

The equation 3h²t +6ht +3t becomes

3t (h²+2h +1)

Now, the resulting quadratic equation is h²+2h +1

Step 2: Finding factors that can be multiplied to give 1 and whose sum is 2

The factors are: 1 and 1

Step 3: The original equation is rewritten as follows:

h² + h + h + 1

Step 4: Factoring the new equation by grouping using the GCF

Where; the perfect square of h and h is h² and the prime factor is h+1

As a result the equation becomes: h (h+1) + 1(h+1)

Step 5: Final factoring by the distributive property

By the distributive property we know that h (h+1) + 1(h+1) is equivalent to (h+1) (h+1)

However, (h+1) (h+1) can be expressed as: (h+1)²

Step 6: Reintroduction of the GCF

The GCF is 3t

Then, the GCF is reintroduced in the equation to get 3t [(h+1)²]

Thus, the answer to this problem is 3t [(h+1)²]

 

 

 

 

Problem No. 102

 

-3 +2y +21y²

Using Box method of factoring to solve this problem, the solution follows the steps below:

Step 1: A 2×2 box is created

 

Step 2: The first term is placed in the top left corner and the last term is placed in the bottom right corner

21y2
	-3

 

Step 3: Finding factors that when multiplied they give -63y² and whose sum is 2y

The factors are: -7y and 9y

These factors are put in the open boxes as follows:

21y2	-7y
9y	-3

 

Step 4: Factoring the terms in each row and in each column

3y	-1
21y2	-7y	7y
9y	-3	3

 

Step 5: The sum of the factors for the columns and the sum of the factors for the rows are the polynomial’s factors

Thus, the answer to this problem is (3y-1) (7y+3)

 

 

Problem No. 64 on pg. 353

 

3x² +17x +10

Using AC method of factoring, the solution follows the steps below:

Step 1: Determining AC by multiplying A and C terms

In 3x² +17x +10 A = 3, B = 17, C = 10

AC = 3*10 =30

Step 2: Finding factors of AC whose sum equals to B (17)

The factors of AC (30) are as follows:

3 and 10

2 and 15

1 and 30

5 and 6

Therefore, factors of AC (30) whose sum is B (17) are 2 and 15

Step 3: The original equation is rewritten as follows:

3x² +15x + 2x + 10

Step 4: Factoring the new equation by grouping using the GCF

Where; the perfect square of x and 3x is 3x² and the prime factor is x+5

3x(x+5) + 2(x+5)

Step 5: Final factoring by the distributive property

By the distributive property we know that 3x(x+5) + 2(x+5) is equivalent to (3x+2) (x+5)

Thus, the answer to this problem is (3x+2) (x+5)
Reference

Dugopolski, M. (2012). Elementary and intermediate algebra, (4^th ed.). New York, NY: McGraw-Hill Publishing.

 

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