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Factor 8 x 2 30 x 27 .

Find the factors of the first and last terms.

The factors of the first term 'eight x squared' and the last term 'negative twenty seven' are shown. The product of the first and the last term is negative two hundred sixteen x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: negative thirty x.

Thus, the 4 x and 9 are to be multiplied, and 2 x and 3 are to be multiplied.

8 x 2 30 x 27 = ( 4 x ) ( 2 x ) = ( 4 x ) ( 2 x 9 ) = ( 4 x + 3 ) ( 2 x 9 )

c h e c k : ( 4 x + 3 ) ( 2 x - 9 ) = 8 x 2 36 x + 6 x 27 = 8 x 2 30 x 27

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Factor 15 x 2 + 44 x + 32 .

Before we start finding the factors of the first and last terms, notice that the constant term is + 32 . Since the product is positive , the two factors we are looking for must have the same sign. They must both be positive or both be negative. Now the middle term, + 44 x , is preceded by a positive sign. We know that the middle term comes from the sum of the outer and inner products. If these two numbers are to sum to a positive number, they must both be positive themselves. If they were negative, their sum would be negative. Thus, we can conclude that the two factors of + 32 that we are looking for are both positive numbers. This eliminates several factors of 32 and lessens our amount of work.
Factor the first and last terms.

The factors of the first term 'fifteen x squared' and the last term 'thirty two' are shown. The product of the first and the last term is four hundred eighty x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: forty-four x.

After a few trials we see that 5 x and 4 are to be multiplied, and 3 x and 8 are to be multiplied.

15 x 2 + 44 x + 32 = ( 5 x + 8 ) ( 3 x + 4 )

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Factor 18 x 2 56 x + 6 .

We see that each term is even, so we can factor out 2.

2 ( 9 x 2 28 x + 3 )

Notice that the constant term is positive. Thus, we know that the factors of 3 that we are looking for must have the same sign. Since the sign of the middle term is negative, both factors must be negative.
Factor the first and last terms.

The factors of the first term 'nine x squared' and the last term 'three' are shown. The product of the first and the last term is twenty seven x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: negative twenty-eight x.

There are not many combinations to try, and we find that 9 x and 3 are to be multiplied and x and 1 are to be multiplied.

18 x 2 56 x + 6 = 2 ( 9 x 2 28 x + 3 ) = 2 ( 9 x 1 ) ( x 3 )

If we had not factored the 2 out first, we would have gotten the factorization

Eighteen x square minus fifty six x plus six is equal to the product of nine x minus one and two s minus six. Two is common between the terms of  the second factor of the polynomial.

The factorization is not complete since one of the factors may be factored further.

18 x 2 56 x + 6 = ( 9 x 1 ) ( 2 x 6 ) = ( 9 x 1 ) 2 ( x 3 ) = 2 ( 9 x 1 ) ( x 3 ) ( By the commutative property of multiplication . )

The results are the same, but it is much easier to factor a polynomial after all common factors have been factored out first.

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Factor 3 x 2 + x 14 .

There are no common factors. We see that the constant term is negative. Thus, the factors of 14 must have different signs.
Factor the first and last terms.

The factors of the first term 'three x squared' and the last term 'negative fourteen' are shown. The product of the first and the last term is negative forty-two x squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: x.

After a few trials, we see that 3 x and 2 are to be multiplied and x and 7 are to be multiplied.

3 x 2 + x 14 = ( 3 x + 7 ) ( x 2 )

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Factor 8 x 2 26 x y + 15 y 2 .

We see that the constant term is positive and that the middle term is preceded by a minus sign.
Hence, the factors of 15 y 2 that we are looking for must both be negative.
Factor the first and last terms.

The factors of the first term 'eight x squared' and the last term 'fifteen y squared' are shown. The product of the first and the last term is one hundred twenty seven x squared y squared. One of the combinations of the factors of the first and the last term yields two new factors of the product such that their sum is the middle term: negative twenty-six xy.

After a few trials, we see that 4 x and 5 y are to be multiplied and 2 x and 3 y are to be multiplied.

8 x 2 26 x y + 15 y 2 = ( 4 x 3 y ) ( 2 x 5 y )

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Practice set a

Factor the following, if possible.

2 x 2 + 13 x 7

( 2 x 1 ) ( x + 7 )

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3 x 2 + x 4

( 3 x + 4 ) ( x 1 )

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4 a 2 25 a 21

( 4 a + 3 ) ( a 7 )

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16 b 2 22 b 3

( 8 b + 1 ) ( 2 b 3 )

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10 y 2 19 y 15

( 5 y + 3 ) ( 2 y 5 )

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6 m 3 + 40 m 2 14 m

2 m ( 3 m 1 ) ( m + 7 )

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14 p 2 + 31 p q 10 q 2

( 7 p 2 q ) ( 2 p + 5 q )

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24 w 2 z 2 + 14 w z 3 2 z 4

2 z 2 ( 4 w z ) ( 3 w z )

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3 x 2 + 6 x y + 2 y 2

not factorable

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As you get more practice factoring these types of polynomials you become faster at picking the proper combinations. It takes a lot of practice!
There is a shortcut that may help in picking the proper combinations. This process does not always work, but it seems to hold true in many cases. After you have factored the first and last terms and are beginning to look for the proper combinations, start with the intermediate factors and not the extreme ones.

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research.net
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there is no specific books for beginners but there is book called principle of nanotechnology
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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
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s. Reply
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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