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How to solve these polynomial problems?

How to solve these polynomial problems? Topic: How to solve these polynomial problems?
July 18, 2019 / By Franklyn
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?
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Best Answers: How to solve these polynomial problems?

Deandre Deandre | 8 days ago
(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
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Deandre Originally Answered: Repost:Can God solve NP-complete problems in polynomial time?
And the angel which I saw stand upon the sea and upon the earth lifted up his hand to heaven, And sware by him that liveth for ever and ever, who created heaven, and the things that therein are, and the earth, and the things that therein are, and the sea, and the things which are therein, that there should be time no longer: Rev 10:5,6 There you go. Time no longer, no longer a prob. The rock question itself needs to be questioned since it violates the logical principle known as the law of the excluded middle which states that a ≠ 'not a'. A shortened form of the question might be phrased, "Can God can't?" Well, can He? Can God actually can not? So when stripped down to its core, the question, can God make a rock so big that even He couldn't move it, is the absurdity, not the concept of omnipotence.

Barry Barry
(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.
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Yvonne Yvonne
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.
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Shelagh Shelagh
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
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Shelagh Originally Answered: How do I solve this polynomial problem?
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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