Answer :
Both terms are multiply of 3, so you can factor 3 out;
3(x² - 9)
Now, we can see that x² - 9 is difference of two perfect squares
Recall that a² - b² = (a+b)(a-b)
So we have x²-9 = (x+3)(x-3)
So we can factor more here; 3(x²-9) = 3(x+3)(x-3)
Final answer: 3(x+3)(x-3)
3(x² - 9)
Now, we can see that x² - 9 is difference of two perfect squares
Recall that a² - b² = (a+b)(a-b)
So we have x²-9 = (x+3)(x-3)
So we can factor more here; 3(x²-9) = 3(x+3)(x-3)
Final answer: 3(x+3)(x-3)
First, factor out a 3.
3(x² - 9)
In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.
In this case, we have x² + 0x - 9. (the 0x is a placeholder)
We want two numbers that add to 0 and multiply to get -9.
Obviously, these numbers are 3 and -3.
Now we have 3(x² + 3x - 3x - 9).
Let's factor.
3(x(x+3)-3(x+3))
3(x-3)(x+3)
There are multiple shortcuts which you could make here, FYI:
Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number)(x+other number).
Whenever you have a difference of squares, like a²-b², that factors to (a+b)(a-b).
3(x² - 9)
In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.
In this case, we have x² + 0x - 9. (the 0x is a placeholder)
We want two numbers that add to 0 and multiply to get -9.
Obviously, these numbers are 3 and -3.
Now we have 3(x² + 3x - 3x - 9).
Let's factor.
3(x(x+3)-3(x+3))
3(x-3)(x+3)
There are multiple shortcuts which you could make here, FYI:
Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number)(x+other number).
Whenever you have a difference of squares, like a²-b², that factors to (a+b)(a-b).
