Question

factorise of z²-6z+9​

Answer 1

Step-by-step explanation:

z²-6z+9

z²-(3+3)z+9

z²-3z+3z+9

(z²-3z)(+3z-9)

z(z-3)+3(z-3)

(z+3)(z-3)

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Answer 2

Question:

Factorise the given equation $z^2-6z+9$.

Answer:

Factorisation : $(z-3)(z-3)$

After using the fromula : $(z^2-3^2)= (z-3)(z+3)$

Explanation:

Here is the equation $z^2-6z+9$

Now factorisng the equation

We can write -6z as -3z and -3z

⇒$z^2-3z-3z+9$

We can take z as common in first and second digit.

We can take -3 as common in Third and fourth digit.

⇒$z(z-3)-3(z-3)$

⇒$(z-3)(z-3)$

Hence we have factorised the equation that $(z^2-3^2)$.

If you want go ahead then

You can see a formula in the eqaution that is$(a^2-b^2)=(a-b)(a+b)$

So Now by using this formula you can get the answer

$(z^2-3^2)= (z-3)(z+3)$

Hence your answer is $(z-3)(z+3)$.