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Usually, the easiest way to do this is to find a pair of numbers whose sum is equal to the coefficient of the first term, and whose product is equal to the product of the coefficients of the first and last terms:

a + b = -3

a * b = -10

take all possible pairs of numbers that multiply to make negative ten:

1 * -10 = -10

-1 * 10 = -10

2 * -5 = -10

-2 * 5 = -10

and then find which one of those pairs also adds up to negative three:

1 + (-10) = -9

-1 + 10 = 9

2 + (-5) = -3

-2 + 5 = 3

So 2 and -5 are the coefficients we want. Now take our original equation:

10a2 - 3a - 1

and separate the middle term into two terms using those coefficients:

10a2 - 5a + 2a - 1

You can then factor a common term out of the first pair and last pair of terms:

5a(2a - 1) + 1(2a - 1)

and then group your coefficients:

(5a + 1)(2a - 1)

Giving you the answer.

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āˆ™ 14y ago
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āˆ™ 13y ago

We have the algebraic expression: 8a2 - 2a

In order to factorize it, we can just simply pull out 2a and we are now left with this: 2a(4a-1)

But, why 2a?

In the given algebraic expression we have two terms: 8a2 and -2a

Now, we have to look for the common factor of 8a2 and -2a.

8a2 = 8 x a x a = 2 x 2 x 2 x a x a

-2a = -2 x a

Common factor is 2a. (2 is also the common factor but we need greatest common factor which is 2a)

4a - 1 can not be further factorized so the final answer is 2a(4a-1).

Visit the link provided below to learn more about factoring.

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āˆ™ 12y ago

Assuming the missing signs are pluses, that factors to (3a + b)(a + 3b)

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āˆ™ 9y ago

(2a - 1)(a - 5)

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āˆ™ 12y ago

(3a - 4)(4a + 3)

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āˆ™ 9y ago

(x + 5)(x - 2)

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āˆ™ 13y ago

2a2 - 11a + 5 = (2a -1) (a - 5)

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āˆ™ 9y ago

2a(4a - 1)

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āˆ™ 12y ago

7a(2a - 3)

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āˆ™ 9y ago

2(a + 5)

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Q: How fo you factor 14a2-21a?
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