Question

factorise
x² + 6x + 9 - y² ​

Answer 1

Step-by-step explanation:

P(x)=x³-2x²-x+2

p(0)= 0 - 0 - 0 + 2 not equal to 0.

p(1)=1 - 2 - 1 + 2=0. x=1 , x - 1 =0.

so, (x- 1)is a factor of p(x) x-1÷x²+6x+9-y² = x²-x-2.

p(x)=(x²-x-2)

=x²-x-2-2

=x(x+1) -2(x+1)

=(x+1) (x-2)

so, x³-2x²-x+2=(x-1)(x-2)(x+1)

• sorry I cannot give exactly answer of the question but I can give a same example

Answer 2

Answer:

$\large\boxed{(x+3+y)(x+3-y)}$

Step-by-step explanation:

To solve this problem, first you can also use perfect square and difference between from two squares.

Given:

First, you rewrite the factor term of x²+6x+9.

9=3²

=x²+6x+3²

Rewrite 6x of 2x*3.

=x²+2x*3+3²

x²+2x*3+3²

$\large\boxed{\textnormal{Perfect Square Formula}}$

$\displaystyle(a+b)^2=a^2+2ab+b^2$

A=X

B=3

$=(X+3)^2$

$=(X+3)^2-y^2$

Solution:

Factor the term of (x+3)²-y²

$\large\boxed{\textnormal{Difference between of two squares formula}}$

x²-y²=(x+y)(x-y)

$(x+3)^2-y^2=((x+3)+y)((x+3)-y)$

Solve

$(x+3)^2-y^2=((x+3)+y)((x+3)-y)=\boxed{(x+y+3)(x+3-y)}$

Therefore, the correct answer is (x+y+3)(x+3-y).