# What is the Greatest Common Factor of 12a and 9a2

What is the Greatest Common Factor of 12a and 9a2

## FACTORING Past GCF

Recollect that the Distributive Property states that

ab + ac  =  a(b + c)

The Distributive Holding allows you to factor out the GCF of the terms in a polynomial to write a factored form of the polynomial.

A polynomial is in its factored class when it is written equally a product of monomials and polynomials that cannot be factored further. The expression 2(3x – 4x) is not fully factored considering the terms in the parentheses have a common factor of x.

## Factoring by Using the GCF

Example 1 :

4xtwo
– 3x

Solution :

Find the GCF :

4x2  =  two
⋅ 2

x

ten

3x  =  3

x

The GCF of 4x2
and 3x is x.

Write terms as products using the GCF as
a factor.

=  4x(x) – 3(x)

Employ the Distributive Property to factor out the GCF.

=
ten(4x

– three
)

Check :

10(4x – 3)  =  x(4x) – x(three)

=
4x2 – 3x

The product is the original polynomial.

Case ii :

10y3 + 20ytwo
– 5y

Solution :

Discover the GCF :

10yiii
=  2

5

y

⋅ y
⋅ y

20y2
=  2
⋅ 2

v

y

⋅ y

5y
=

5

y

The GCF of

10y
3
, 20y
2
and 5y is 5y.

Write terms as products using the GCF as
a cistron.

10y3 + 20y2 – 5y  =  2y2(5y) + 4y(5y) – ane(5y)

Employ the Distributive Property to factor out the GCF.

=5y(2ytwo + 4y – 1)

Check :

5y(2yii + 4y – 1)  =5y(2y2) + 5y(4y) + 5y(-1)

=  10ythree + 20y2 – 5y

The production is the original polynomial.

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Example 3 :

-12a – 8aii

Solution :

Both coefficients are negative. Factor out -1.

-12a – 8a
2
=-i(12a + 8a
2)

Find the GCF :

12a  =
2

2

iii

a

8a2  =
2

2

⋅ 2

a

⋅ a

The GCF of

12a and 8a
2
is 4a.

Write terms as products using the GCF as
a factor.

=  -ane[3(4a) + 2a(4a)]

Use the Distributive Property to factor out the GCF.

=  -1[4a(three + 2a)]

=  -4a(3 + 2a)

Check :

-4a(3 + 2a)  =  -4a(three) – 4a(2a)

=  -12a – 8atwo

The product is the original polynomial.

Example four :

5x2
+ 7

Solution :

Discover the GCF :

5x2
=

5

x

⋅ x

seven  =  seven

At that place are no mutual factors other than one.

The polynomial cannot be factored.

Sometimes the GCF of terms is a binomial. This GCF is called a common binomial factor. You factor out a common binomial factor the same fashion you lot cistron out a monomial gene.

## Factoring Out a Common Binomial Gene

Factor each expression.

Example v :

vii(x – 3) – 2x(10 – 3)

Solution :

(x – three) is a common binomial gene.

=  7(x – 3)
– 2x(x – three)

Cistron out (x – iii).

=
(x – iii)(7 – 2x)

Example 6 :

-y(y2 + 5) + (y2 + five)

Solution :

(y
2
+ five) is a common binomial factor.

=  -y(y2 + 5)
+
(yii + five)

=  -y(y2 + 5) + 1(ytwo + 5)

Factor out (y
2
+ 5).

=(
y2 + 5
)(-y + ane)

=(
y2 + 5
)(1 – y)

Example 7 :

9n(n + iv) – v(4 + n)

Solution :

4 + due north  =  n + 4

So,

=  9n(due north + 4) – five(n + 4)

(n + four) is a mutual binomial factor.

=  9n(n + 4)
– five(n + 4)

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Factor out (n + 4).

=(n + four)(9n – 5)

Example 8 :

-3y2(y + 2) + 4(y – vii)

Solution :

There are no common factors.

The expression cannot be factored.

Y’all may be able to factor a polynomial by grouping. When a polynomial has iv terms, y’all can make two groups and gene out the GCF from each grouping.

## Factoring by Group

Example 9 :

12x3 – 9xtwo
+ 20x – fifteen

Solution :

Grouping terms that have a common number or variable every bit a factor.

=  (12x
3
– 9x
2) + (20x – 15)

Cistron out the GCF of each group.

=
3xtwo
(4x – iii) +
5(4x – iii)

(4x – 3) is a common factor.

=  3x2
(4x – 3)
+ 5(4x – 3)

Factor out (4x – three).

=
(4x – 3)(

3xii + 5)

Check :

(4x – 3)(
3xii + 5)  =  4x(3x2) + 4x(5) – 3(3xtwo) – 3(5)

=  12x3 + 20x – 9xii – 15

=
12x3 – 9x2 + 20x – xv

The product is the original polynomial.

Case 10 :

9a3 + 18a2 + a + ii

Solution :

Group terms that take a common number or variable every bit a factor.

=  (9a3 + 18a2) + (a + 2)

Gene out the GCF of each grouping.

=
9a

2
(a + 2) +
1(a + two)

(a + 2) is a mutual gene.

=  9a2
(a + 2) + 1(a + 2)

Gene out (a + 2).

=(a + 2)(9a

2 + ane)

Bank check :

(a + 2)
(9a

two + one)  =  a(9a2) + a(1) + two(9aii) + 2(1)

=  9a3 + a + 18a2 + two

=  9a3 + 18a2 + a + 2

The product is the original polynomial.

Recognizing opposite binomials can help you factor polynomials. The binomials (5 – x) and (x – 5) are opposites. Notice (v – x) can be written as -one(x – 5).

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=
-1(x – v)

Distributive Property.

=
(-1)(x) +(-i)(-5)

Simplify.

=  -10 + v

Commutative Holding of Improver.

=  5 – x

So,

5 – x  =  -1(ten – five)

## Factoring with Opposites

Instance 11 :

Factor 3x3
– 15x2
+ 10 – 2x past group.

Solution :

=  3x3 – 15x2 + ten – 2x

Group terms.

=  (3x3 – 15xii) + (10 – 2x)

Factor out the GCF of each group.

=  3xtwo(ten
– v
) + 2(five – 10)

Write (five – 10) as -1(10 – 5).

=  3x2(10 – five) + 2(-i)(10 – five)

=  3x2(x – 5) – 2(x – v)

(x – 5) is a common cistron.

=  3x2
(x – 5)
– 2(x – v)

Factor out (10 – 5).

=
(x – v)(3x

2 – 2)

## Scientific discipline Awarding

Case 12 :

Lily’southward calculator is powered by solar energy. The area of the solar panel is (7x2
+ x) cm2. Factor this polynomial to find possible expressions for the dimensions of the solar console.

Solution :

A  =7x
2
+ 10

The GCF of 7x2
and ten is ten.

Write each term as a product using the GCF as a factor.

=  7x(x) + ane(x)

Use the Distributive Property to factor out the GCF.

=  10(7x + 1)

Possible expressions for the dimensions of the solar panel are 10 cm and (7x + 1) cm.

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