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Serga [27]
3 years ago
5

Celina says that each of the following expressions is actually a binomial in disguise:(Expressions in picture) For example, she

sees that the expression in (i) it is algebraically equivalent to − , which is indeed a binomial. (She is happy to write this as + (−), if you prefer.) Is she right about the remaining four expressions?

Mathematics
1 answer:
Juliette [100K]3 years ago
4 0

Yes , she is right. All the remaining can be expressed in binomial.

II) 5x^3 * 2x^2 - 10x^4 + 3x^5 + 3x * (-2)x^4

10x^5 - 10x^4 + 3x^5 + -6x^5= 7x^5 - 10x^4 is a binomial

iii) (t+2)^2 - 4t = t^2 + 4t + 4 -4t = t^2 + 4 is a binomial

iv) 5(a-1) - 10(a-1) + 100 ( a-1) = 5a - 5 - 10a + 10 + 100a -100 = 95a -95 is a binomial

v) ( 2\pi  r -\pi  r^2)r - ( 2\pi  r -\pi  r^2)2r = 2\pi  r^2 -\pi  r^3 - 4\pi  r^2 + 2\pi  r^3 = -2\pi  r^2  + \pi  r^3 is also a binomial



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(f/g)(x) = [ f(x) ]/[ g(x) ]
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One way to do this is to notice

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