Examples of divisor in the following topics:
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- For example, find the quotient and the remainder of the division of $x^3 - 12x^2 -42$, the dividend, by $x-3$, the divisor.
- Multiply the divisor by the result just obtained (the first term of the eventual quotient): $x^2 \cdot (x − 3) = x^3 − 3x^2$.
- For example, find the quotient and the remainder of the division of $x^3 - 12x^2 -42$, the dividend, by $x-3$, the divisor.
- Divide the first term of the dividend by the highest term of the divisor (meaning the one with the highest power of $x$, which in this case is $x$): $x^3 \div x = x^2$.
- Multiply the divisor by the result just obtained (the first term of the eventual quotient): $x^2 \cdot (x − 3) = x^3 − 3x^2$.
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- Polynomial long division functions similarly to long division, and if the division leaves no remainder, then the divisor is called a factor.
- So we write down a $3x^2$, multiply the divisor with this result and subtract this from the dividend:
- As multiplying any polynomial with the divisor $2x-4$ gets us a polynomial of degree greater than $0$, we cannot divide any further.
- This means that $D(x)=d(x)q(x)$: the dividend is a multiple of the divisor, or the divisor is said to $$divide the dividend.
- We say that the divisor is a factor of the dividend.
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- It states that the remainder of a polynomial f(x) divided by a linear divisor (x-a) is equal to f(a).
- We start with writing down the coefficients from the dividend and the negative second coefficient of the divisor.
- As the leading coefficient of the divisor is $1$, the leading coefficient of the quotient is the same as that of the dividend:
- The result of $-12 + 3$ is $9$, so since the leading coefficient of the divisor is still $1$, the second coefficient of the quotient is $-9:$
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- If $a_0$ and $a_n$ are nonzero, then each rational solution $x=p/q$, where $p$ and $q$are coprime integers (i.e. their greatest common divisor is $1$), satisfies:
- Since any integer has only a finite number of divisors, the rational root theorem provides us with a finite number of candidates for rational roots.
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- This can be solved using the property that if $x_0$ is a zero of a polynomial, then $(x-x_0)$ is a divisor of this polynomial and vice versa.
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- If the dividend and the divisor have the same sign, that is to say, the result is always positive.
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- Therefore, we use the cancellation method to simplify the numbers as much as possible, and then we multiply by the simplified reciprocal of the divisor, or denominator, fraction:
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- Recall the rule for dividing fractions: the dividend is multiplied by the reciprocal of the divisor.