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Factorise fully 64(𝑥 + 1)² − 9(𝑥 − 1)².
Factorize fully 64 times 𝑥 plus one squared minus nine times 𝑥 minus one squared.
Now we could factorize this by multiplying everything out. Essentially, expanding it and then simplifying it, putting it back together, and then factoring it from there. But actually, there is something much faster. If we would let this be our 𝑎 term and this be our 𝑏 term and there’s a minus in between, we’re very close to the difference of squares form. So difference of square states: if we have 𝑎 squared minus 𝑏 squared, it can factor to equal 𝑎 plus 𝑏 times 𝑎 minus 𝑏.
So looking at our first term, could we let that be 𝑎 squared? Can 64 and 𝑥 plus one squared be square rooted? Because if we would let this be 𝑎 squared, in order to find 𝑎, we would need to square root that. The square root of 64 is eight. And the square root of 𝑥 plus one squared would be 𝑥 plus one. So it works! And same for the second. Can that be represented by 𝑏 squared? It can, because the square root of nine is three and the square root of 𝑥 minus one squared is 𝑥 minus one. So we have an 𝑎 squared. We have a 𝑏 squared. And we have the minus sign between them. So this will be considered a difference of squares.
So now we need to simply take 𝑎 and 𝑏 and plug them in to the factorize form. Just to keep this organized, while we’re plugging in, we’ll let the 𝑎 be pink and the 𝑏 be blue. So here, we’ve plugged in 𝑎 plus 𝑏. And now, we’ve plugged in 𝑎 minus 𝑏. So in order to completely factor, we need to simplify inside the parentheses. And to do so, we will need to distribute. So we’ve distributed most of it. But we have to be careful of this last piece because it is minus 𝑏. So if we would distribute the three, it creates three 𝑥 minus three. However, we need to distribute the negative sign in front of that. So really that’s a negative three 𝑥 and a positive three.
Now we need to combine like terms in the parentheses. We get 11𝑥 plus five for the first one and five 𝑥 plus 11 on the second one. So our final answer will be 11𝑥 plus five times five 𝑥 plus 11. Now, I stated before there is another way to do this. It would take a little bit more work. But we can do that as well. So let’s go ahead and erase all of this, but keep our final answer and show that there’s another way to do this.
Okay, so starting over, we will factorize, but using a different method. We will begin by expanding this, multiplying it out. First, we will take care for the squares. So when something is squared, we’re multiplying it to itself. So instead of 𝑥 plus one squared, it’s 𝑥 plus one times 𝑥 plus one and the same thing with 𝑥 minus one. And now, we need to FOIL the 𝑥 plus one times 𝑥 plus one. We’ll distribute, and we get 𝑥 squared plus 𝑥 plus 𝑥 plus one. So we can combine the two 𝑥s in the middle and make it 𝑥 squared plus two 𝑥 plus one. And now we do the same for the 𝑥 minus one times 𝑥 minus one. And we get 𝑥 squared minus one 𝑥 minus one 𝑥 plus one. So we can combine the middle two terms and get 𝑥 squared minus two 𝑥 plus one.
So now, we distribute the 64 and the negative nine. Now that we’ve distributed, we need to combine like terms. And we get 55𝑥 squared plus 146𝑥 plus 55. So this is what we would call an advance trinomial because 𝑎 is greater than one. 𝑎 is the leading coefficient, 55. So to solve, we will use the slip and slide method. So we will slip the 55 to the back and take 55 times 55. And we get 𝑥 squared plus 146𝑥 plus 3025. So two numbers that multiply to be 3025 and add to be 146, would be 25 and 121. Now, the 55 that we slipped to the back, now we must slide underneath these numbers.
And now we simplify our fractions. And we get 𝑥 plus five 11ths times 𝑥 plus 11 fifths. Now, our denominators we don’t leave on the denominator, the bottom, we will move them up with the 𝑥s. So just as we got before, the final answer is 11𝑥 plus five times five 𝑥 plus 11.
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