Question

(x^2+ 6x + 9) / (x-3)

Answer 1

Answer:

Simplify:

$\frac{( {x}^{2} + 6x + 9) }{x - 3}$

Factoring:

${x}^{2} + 6x + 9$

The first term is, x2 its coefficient is 1 .

The middle term is, +6x its coefficient is 6 .

The last term, "the constant", is +9

Step-1 : Multiply the coefficient of the first term by the constant 1 • 9 = 9

Step-2 : Find two factors of 9 whose sum equals the coefficient of the middle term, which is 6 .

-9 + -1 = -10

-3 + -3 = -6

-1 + -9 = -10

1 + 9 = 10

3 + 3 = 6 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 3 and 3

${x}^{2} + 3x + 3x + 9$

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x+3)

Add up the last 2 terms, pulling out common factors :

3 • (x+3)

Step-5 : Add up the four terms of step 4 :

(x+3) • (x+3)

Which is the desired factorization.

Multiply (x+3) by (x+3)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (x+3) and the exponents are :

1 , as (x+3) is the same number as $(x+3) {}^{1}$

and 1 , as (x+3) is the same number as$(x + 3) {}^{1}$

The product is therefore, $(x + 3) {}^{(1 + 1)} = (x + 3) {}^{2}$