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mcshweezy

  • 2 years ago

What is the remainder when (3x4 + 2x3 – x2 + 2x – 14) ÷ (x + 2) ?

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  1. hoblos
    • 2 years ago
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    10

  2. JamesJ
    • 2 years ago
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    @hoblos: if you answer it, you explain it!

  3. mathmate
    • 2 years ago
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    What you do is to evaluate the numerator, using x=-2 (from x+2=0), and what you get is the remainder, namely 10 as Hoblos posted.

  4. cwrw238
    • 2 years ago
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    this is called the Remainder THEorem

  5. jhonyy9
    • 2 years ago
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    (3x4+2x3-x2+2x-14) : (x+2) =3x3-4x2+7x-12 3x4+6x3 --------- 0 -4x3-x2 -4x3-8x2 ---------- 0 7x2+2x 7x2+14x ---------- 0 -12x-14 -12x-24 --------- 0 10

  6. JamesJ
    • 2 years ago
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    @cwrw: right. The idea is that given any polynomial p(x) and a linear term such as (x+2), then we write p(x) as p(x) = (x+2)q(x) + r --- (*) where q(x) is another polynomial. r is the remainder after p(x) is divided by (x+2). ===== If x=-2 is a root of the polynomial p(x), then p(-2) = 0. Hence using equation (*), 0 = p(-2) = (-2+2)q(-2) + r = 0 + r i.e., r = 0. If x=-2 is not a root of p(x), then r will not be zero. It can be calculated by evaluating p(-2), because p(-2) = (-2+2)q(-2) + r = 0 + r i.e., r = p(-2) ===== So this is why a short-cut way to finding the remainder of p(x) = 3x^4 + 2x^3 – x^2 + 2x – 14 when divided by (x+2) is just to evaluate p(x) for x = -2: r = p(-2) = 3(-2)^4 + 2(-2)^3 - (-2)^2 + 2(-2) - 14 = 48 - 16 - 4 - 4 - 14 = 10

  7. malice
    • 8 days ago
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    The answer is *not* ten, just took the test

  8. mathmate
    • 7 days ago
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    @malice If the answer is *not* ten, it could be one of the many possibilities: 1. the test has an incorrect answer 2. the test question is different from what was posted above (typo, or the test is automorphic, i.e. changes the data every time the question is displayed) 3. All three or four people who responded to the question made the same mistake with probability < \(0.01^4=10^{-8}\). Actually this is an excellent lesson to learn: We do not take answers from someone else and use it as our own work (plagiarism). Work out the problem personally so that we know HOW to do the work. Finally, you also realize that getting \(help\) or answers from this site or anywhere else for an online test is considered cheating, and is taken very seriously by the authorities.

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