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Factoring: Special Cases Part I  Problem 1
Alissa Fong
Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
When asked to factor this problem, a lot of students recognize there is a number that multiplies into both terms. A lot of students look at this, and they can already see that the greatest common factor is going to be 8.
So what most students will do is write this down, and they'll say I'm done. You guys this problem can actually be factored further. Look here.
This is what we call the difference of perfect squares. I can rewrite this as x minus 4, times x plus 4. The reason why is, because difference of perfect squares is when you see something that looks like this, two things that are squared like we had x² and 4², then you can write it out in factored form; like a minus b, a plus b.
So be careful when you're doing your factoring problems. A lot of times when students see a binomial that has two terms, they think they're done. But sometimes binomials can be factored even further into two more binomials.
It's really tricky. You'll get the hang of it when you do more practice, but this is the final answer. This is the most completely factored form of that original statement.
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Alissa Fong
M.A. in Secondary Mathematics, Stanford University
B.S., Stanford University
Alissa has a quirky sense of humor and a relatable personality that make it easy for students to pay attention and understand the material. She has all the math tips and tricks students are looking for.
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Factoring: Special Cases Part I
Problem 1 7,219 viewsFactor:
8x² − 32 
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Problem 2 6,242 viewsIf the area of a square is 49m² + 28m + 4, find the side length.

Factoring: Special Cases Part I
Problem 3 5,763 viewsFind the missing term of the perfect square trinomial:
9x² − ___ + 25 
Factoring: Special Cases Part I
Problem 4 1,222 views 
Factoring: Special Cases Part I
Problem 5 1,218 views 
Factoring: Special Cases Part I
Problem 6 828 views
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