What best describes a neutral solid object that is charged by conduction?
Am guessing this is not high school work
Answer:
2x(x+y)(x-y)
Step-by-step explanation:
Given the expression;
(x+y)² + 2(x+ y)(x- y) + (x– y)²
Factorize
(x+y)² + (x+ y)(x- y) + (x+ y)(x- y) + (x– y)²
= (x+y)[x+y+(x-y)] +(x-y)[(x+y)+(x-y)]
= (x+y)(x-y)(x+y+(x-y))
= (x+y)(x-y)(x+y+x-y)
= (x+y)(x-y)(2x)
hence the expanded form is 2x(x+y)(x-y)
Answer:
Step-by-step explanation:
You need to know this indice rule:
If you multiply two terms with the same base, then add their powers and multiply their coefficients.
: Here we have to multiply each term in the first pair of brackets by the term in the second pair of brackets.
So we can split this into parts:
Now using the rule above expand the brackets:
: Notice how 
since 
so 
.
: Be careful with negatives, two negatives make a positive.
Now add them all together:
Answer:
What equation? Here’s a big tip for you, equations contain an ‘equals sign’ (‘=’) and something that they are equal to. I’m assuming your equation is:
ax2+bx+c=0
I’m going to use p and q for the roots of the equation as they are easier to type than α and β . Thus we have:
a(x−p)(z−q)=0⇒ax2−a(p+q)x+apq=0
Equating coefficients with our initial equation:
x1 : b=−a(p+q)⇒p+q=−ba
x0 : c=apq⇒pq=ca
Now p2+q2=p2+2pq+q2−2pq=(p+q)2−2pq
=(−ba)2−2ca
=b2a2−2ca
=b2−2aca2
