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Topic: How to solve these polynomial problems?**Question:**
(x-3y)^2
(3b-c)^3
i'm having issues understanding how to do these :/
could you explain the second one? how you multiplied it?

July 18, 2019 / By Franklyn

(x-3y)^2 (x-3y)(x-3y) x^2-3xy-3xy+9y^2 x^2-6xy+9y^2 (3b-c)^3 (3b-c)(3b-c)(3b-c) (9b^2-3bc-3bc+c^2)(3b-c) 27b^3-9b^2c-18b^2c+6bc^2+3bc^2-c^3 27b^3-27b^2c+9bc^2-c^3

👍 264 | 👎 8

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(x - y)^2 means to multiply the binomial by itself twice. That is (x - y)*(x - y). If you use the FOIL (IE first, outer, inner, last) process, this produces x^2 - xy - xy + y^2. Adding like terms, you end up with x^2 - 2xy + y^2 (3b - c)^3 means to multiply the binomial by itself three times.So using FOIL, as above, for (3b - c)*(3b - c) will give you 9b^2 -6bc +c^2. Now multiply this trinomial by the binomial for your answer. Here you will have to do it term by term. So take 3b and multiply times each term to get 27b^3 - 18b^2c + 3bc^2; then multiply the trinomial by (-c) to get -9b^2c + 6bc^2 - c^3.. Finally add the like terms in the two products to get 27b^3 - 27b^c +9bc^2 - c^3.

👍 110 | 👎 7

You're actually not solving then, but expanding them. . . Write them out in full first, so you get: (x-3y)(x-3y) and (3b-c)(3b-c)(3b-c) and then equate them using the FOIL rule, "first, outside, inside, last"; that is you multiply out the components of the brackets in that order and then add the products together. . . so for the first one you get: x*x=x^2, x*-3y= -3xy, x*-3y=-3xy, and -3y*-3y=9y^2 then you add all these together; x^2-3xy-3xy+9y^2, which gives x^2-6xy+9y^2. For the cubic you expand out one pair of the brackets the same, and then multiply that expansion by the last set of brackets by multiplyng first the first part of the bracket by everything and then the second part of the racket by everything and again adding it all together. Hopefully this makes sense.

👍 103 | 👎 6

do you want us to expand these? (X-3y)^2=x^2-6xy+9y^2 (3b-c)^3 = (3b)^3 -3(3b)^2(c)+3(3b)(-c)^2 -c^3 =27b^3-27b^2c+9bc^2-c^3

👍 96 | 👎 5

As per remainder theorem, when f(x) is divided by x - a, the remainder is f(a) So when P(x) = x^3 - x - 2 is divided by x - 3, the remainder is P(3) P(3) = 3^3 - 3 - 2 = 27 - 5 = 22

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