Polynomials and the Binomial Theorem

The Binomial Theorem #

Pascal triangle hands you the coefficients.

The rows of Pascal’s triangle are the coefficients of an expanded bracket

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Squaring a bracket gives coefficients 1, 2, 1.

Cube it: multiply the square by one more bracket and collect the like terms.

Cubing gives 1, 3, 3, 1. Each number is the two above it added.

Why they add: $a^{2}b$a²b comes from $a^2$a² times b and from 2ab times a, so its coefficient is 1 + 2 = 3.

So the triangle gives you every expansion without multiplying the brackets out.

Each entry counts choices: in row 4, the 6 counts the ways to pick 2 items from 4.

The general term is $nCr{a}^{n-r}{b}^r$nCr aⁿ⁻ʳ bʳ, where nCr counts the ways to choose which r brackets give a b.

Binomial Series for a Negative or Fractional Index #

It never terminates, and it needs |x| < 1.

The binomial expansion runs forever whenever the power is not a positive whole number

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When n is a positive whole number one factor becomes zero, and that is why the expansion stops.

Put n = −1 in the same formula and nothing ever vanishes: the series runs forever.

Near zero the four-term series follows $\frac{1}{1+x}$1/(1 + x) closely, but past x = 1 it moves away.

Any value of n uses the same formula, and all of them need |x| < 1.

Expanding (a + bx) to a Rational Power #

Factor the constant out, and the range moves too.

Taking the constant out of the bracket turns any binomial power into the standard series

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The series works only on brackets that start with 1, so take the constant out first.

Expand as usual with $\frac{x}{4}$x/4 wherever the formula says x, then multiply the 2 back in.

The condition moves with the substitution: the range here is |x| < 4, not |x| < 1.

In general the range is set by the ratio of the two numbers in the bracket.

Factor and Remainder Theorem #

Substitute, and you have the remainder.

Substituting a value into a polynomial tells you the remainder when you divide by that bracket

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Dividing 17 by 5 gives 3 remainder 2, which is the same fact as 17 = 5 × 3 + 2.

Polynomials divide the same way: f(x) = (x − a) q(x) + r, where r is a constant.

Now substitute x = a into that identity. The bracket a − a is equal to zero.

Zero times q(a) is zero, so that term disappears and f(a) is the remainder r.

Try it: divide $f\left(x\right)=x^2+5$f(x) = x² + 5 by (x − 3). Then f(3) = 9 + 5 = 14, the remainder.

Check it: (x − 3)(x + 3) is $x^2-9$x² − 9, and adding 14 gives back $x^2+5$x² + 5.

If f(a) comes out as zero, there is no remainder and the bracket is a factor.

A factor gives a root, so the curve crosses the x-axis at that value of x.

Solving a Cubic Completely #

Hunt one root, divide, finish on the quadratic.

Find one root among the divisors of the constant, divide it out, and finish with the quadratic

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Look for a root among the divisors of the constant term: f(1) is 0, so x − 1 is a factor.

Divide the cubic by that factor, by long division or by matching coefficients. A quadratic is left.

Factor the quadratic and the cubic is done: the roots are 1, 3 and −2.

Roots and Coefficients of a Cubic #

Three sums, read off without solving.

The three symmetric sums of the roots of a cubic are read straight off its coefficients

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A cubic with roots $\alpha$α, $\beta$β and $\gamma$γ is a times three brackets. Expand it and look at the pattern.

Match the coefficients one by one and three sums appear, with the signs alternating.

All three are read off the coefficients, without finding a single root.

A quartic follows the same pattern with one more sum, and the signs keep alternating.

Square $\sum \alpha$Σα and take away twice $\sum \alpha \beta$Σαβ, and you have the sum of the squares.

Transforming the Roots of a Polynomial #

Double every root without finding one.

Substituting the reverse of a transformation builds the equation whose roots are transformed

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Sometimes the roots you need are the old ones doubled, shifted, or turned into reciprocals.

Write the transformation, then rearrange it to give x in terms of y.

Substitute and tidy up. The new equation is the old one written in the new variable.

Check by factoring: the roots really are 4 and 6, which is what was asked for.

The rule is always the same: write x in terms of y, then substitute that in place of x.

Expanding Three Binomials #

Two brackets first, then the third.

Expand two brackets first, then multiply every term of the result by each term of the third

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Expand two of the brackets first. The result has three terms, and the last bracket is still to multiply.

Multiply every term by x, then every term by 3: six products. Then collect the like terms.

The coefficients follow a pattern: the $x^2$x² term adds the three numbers, and the constant multiplies them.

Check by putting in a number: x = 1 gives 24 from the brackets and from the expansion.