Week 6 – Precalculus 11 – Factoring

This week we learned more about factoring, and it is a fundamental step in completing quadratic equations so I thought that I would write my blog post about it.

Say you wanted to factor a^{2}-9.

Think to yourself: if you were to multiply two expressions together, what would yield a^{2}-9? Well, by taking a look at a^{2}, we know that we need two a’s to equal a^{2}. And to yield -9, we know that we need two 3’s to equal -9, but one has to be negative, or we will get a different answer than a^{2}-9.

So, with that knowlege, we now have a+3 & a-3. But when you’re factoring, it’s important to have the brackets around each individual expression so that we know to multiply.

Our final answer should be \left(a+3\right)\left(a-3\right).

Seem simple enough? Let’s move on to a more difficult question.

Say you wanted to factor -2x^{2}+6x+20.

When we look at this expression, ask yourself: is there a greatest common factor? By taking a close look, we notice that there is a greatest common factor in this expression, -2. We want to pull that out and leave it outside of the expression now as it makes life a whole lot easier.

Now we have -2\left(x^{2}-3x-10\right), because -2 multiplied by each number in this expression will equal the original answer unfactored expression.

Because we factored out the GCF, we can see that we can finish this factoring question by factoring the trinomial.

Our final answer will be -2\left(x+2\right)\left(x-5\right) because -5 plus 3 will equal 3, and -5 multiplied by 2 will equal -10.

I hope this helps!

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