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I need to factor each one of these polynomial

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3 Answers

First put the polynomial in standard form (descending order of exponents):

2a^2 + 3a + 1

now factor the first coefficient (2):

2 * 1

Put these factors inside parenthesis:

(2a      )(a     )

Now, factor the last term (1):

1 * 1

These factors will now go inside the parenthesis:

(2a    1)(a    1)

You will end up multiplying the factors, and adding them
together to get the single order term (3a), the resulting sum must equal that
factor. Since the single order term (3a) is positive, the factors, when added
together must be positive.

So, 2a * 1 = 2a, a * 1 = a, Adding the results together, 2a + a = 3a

So, Both factors must be positive:

(2a  +  1)(a  + 1) = 2a^2 + 2a + a + 1,
 
combining like terms:
 
2a^2 + 3a + 1
The second one can be factored using quadratic formula or, if you do not know one, by a simple trick:
 
2a2+3a+1=2a2+2a+a+1=(2a2+2a)+(a+1)=2a*(a+1)+(a+1)=(a+1)*(2a+1)
 
The first one can be easily factored by first taking w out as a common factor.
 
9w-w3=(9-w2)*w;
 
Now 9-w2 can be factored even further by using a2-b2=(a+b)*(a-b) identity.
 
9-w2=32-w2=(3+w)*(3-w);
 
So your final answer for the first one is:
 
9w-w3=w*(w+3)*(3-w)

Hello Glo.

To factor the 1st one, you should find the Greatest Common Factor or GCF. You can use a factor tree if you are used to finding the GCF that way. So,


        9w      -      w^3
        / \               / \
       9   w          w^2 w
      / \ |            / \    |
      3 3 w          w w  w

A common factor will be

w

so divide each term by W:

9w /w = 9, and w^3/ w = w^2

You will be making a distribution problem:

w(9 - w^2) = w(3 +w)(3-w)

 
 
 

Comments

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