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## Algebra 2

### Course: Algebra 2>Unit 3

Lesson 6: Polynomial identities

# Analyzing polynomial identities

Sal considers two cases of polynomial identities and their proofs.

## Want to join the conversation?

• On the second exercise, first choice, couldn't it be true if either x or y was equal to 0? • I understand that the second step: (x-2) (x-2) expands into the form: ( x^2 -4x +4 ).
However, if this part of the equation is in the form (a-b) ^2 , which expands into
(a^2 - 2ab +b^2) , why then isn't the middle term positive? If x= a and -2= b , shouldn't it also equal ( (x) ^2 -2 (x) ( -2) + (-2)^2 ). Here one can see that the statement equals x^2 + 4x +4 . Does this mean that the b term only includes 2 and not -2? Thanks. • They went wrong in the 4th step.When you multiply 2 negative numbers you get a positive number. So -3 x -4x = +12x • If x^2+[1/(x^2)]=72 then find the value of x^3+[1/(x^3)] • Hi, sorry I know this question doesn't exactly relate to the video but it is in the section of my textbook called Quadratic Identities...
I'm stuck on this question:
Express m^2 in the form a(m-1)^2+b(m-2)^2+c(m-3)^2
Would really appreciate some help or otherwise direction to a video which could help.
Thank you!
(1 vote) • They want you to find a, b, and c that satisfy a(m-1)²+b(m-2)²+c(m-3)²=m²
To start, let's expand the left:
a(m²-2m+1)+b(m²-4m+4)+c(m²-6m+9)=m²
am²-2am+a+bm²-4bm+4b+cm²-6cm+9c=m²
Now regroup it:
am²+bm²+cm²-2am-4bm-6cm+a+4b+9c=m²
(a+b+c)m²-2am-4bm-6cm+a+4b+9c=m²

On the left, we have constants, a+4b+9c. On the right, no constants. So a+4b+9c must equal 0.
Similarly, there are no linear (m¹) terms on the right. So -2am-4bm-6cm=0.
So the m² terms must be equal. (a+b+c)m²=m².

Let's factor the second-to-last equation. We get
-2m(a+2b+3c)=0
So either m=0, or a+2b+3c=0. We reject m=0, because we want this to be true for all m. So a+2b+3c=0.

For the last equation, we subtract m² to get (a+b+c)m²-m²=0
m²(a+b+c-1)=0
So a+b+c-1=0, or a+b+c=1

Now we finally have a system of equations:
a+4b+9c=0
a+2b+3c=0
a+b+c=1

I'll skip the algebra to solve the system, but we get solutions a=3, b= -3, c=1

So m²=3(m-1)²-3(m-2)²+(m-3)²
• In the second problem, how did he get 4n by squaring? I tried it myself and I still could not figure it out. Of course the problem couldn't be solved without the 4n, but how did Sal get to that point? • (n+2)^2=(n+2)(n+2)
When you multiply (n+2) and (n+2), you get n^2+2n+2n+4. Simplifying, you get n^2+4n+4 since 2n+2n=4. The 4n comes from the middle terms of the expanded polynomial, 2n and 2n, being added together.
Similarly, if we had (x+5)(x+5), the answer would be x^2+5x+5x+25, or x^2+10x+25. The 10x comes from the middle terms, 5x and 5x, being added together.
Hope this helps!
• Can someone help me with this algebra question?
Prove 3(x+y)^2 - (x-y)^2 = 2x^2 + 8xy^2 +2y^2
Thank you so much!
(1 vote) • I understand that the second step: (x-2) (x-2) expands into the form: ( x^2 -4x +4 ).
However, if this part of the equation is in the form (a-b) ^2 , which expands into
(a^2 - 2ab +b^2) , why then isn't the middle term positive? If x= a and -2= b , shouldn't it also equal ( (x) ^2 -2 (x) ( -2) + (-2)^2 ). Here one can see that the statement equals x^2 + 4x +4 . Does this mean that the b term only includes 2 and not -2? Thanks.
(1 vote) • I'm a little confused about the first example, I thought that (x-2)^2 is equal to (x-2)(x+2), because negative times positive is negative, but negative times negative is positive!? Wouldn't (x-2)(x-2) equal (x+2)^2?
(1 vote) • An exponent is repetitive multiplication of the same value.
For example: 5^2 = 5*5; (-5)^2=(-5)(-5)
The same concept extends to polynomials.
The expression: (x-2)^2 is telling you to multiply the factor (x-2) with itself = (x-2)(x-2)

In your examples, you are trying to change the factor's that the exponent applies to. All your new versions are completely different than the original. You can verify this if you use FOIL to multiply your various options:
(x-2)^2 = (x-2)(x-2) = x^2-2x-2x+4 = x^2-4x+4
(x+2)^2 = (x+2)(x+2) = x^2+2x+2x+4 = x^2+4x+4
(x-2)(x+2) = x^2+2x-2x-4 = x^2-4

Hope this helps.
(1 vote)
• Why (x-2)^2 didn't expand into (x-2) (x+2) in accordance with the formula?
(1 vote) 