start with quadratic equation: ax² + bx + c = 0
Rewrite as x² + (b/a)x = -c/a That becomes (x + b/2a)² = -c/a + (b/2a)² Call d = b/2a: (x + d)² = d² - c/a Take square roots: x + d = ±√(d² - c/a), and simplify (Solving for x gives two answers, call it u and v. The quadratic equation can then be rewritten as: a(x+u)(x+v) = 0)
Just compute the values: equation of square: a(x+d)² + e = 0, where d = b/(2a) and e = c - b²/(4a)
steps to get there: use method 1 to get (x + d)² = d² - c/a multiply both sides by a: a(x + d)² = ad² - c call e = c - ad², then a(x + d)² + e = 0 (x + (b/a))² + c/a - ((b/a)/2)² = 0 \==========/ quadratic + remainder = 0
-b/a is the vertex (x value) if graphing, y = remainder
which leads to quadratic formula: x = (-b +- √(b²-4ac)) / 2a
Start with equation of square.
If a>0 then max = infinity, min is at x = -d If a<0 then min = -infinity, max is at x = 0
Multiplicity is the number of times a given factor appears in the factored form of the equation of a polynomial.
A factor with an even multiplicity will touch, odd multiplicity will cross.
Example: f(x) = (x+3)(x-2)²(x+1)³ x = -3 has multiplicity of 1, crosses the x axis like a line x = 2 has multiplicity of 2, touches the x axis like a parabola x = -1 has multiplicity of 3, crosses the x axis in s shape (half-flipped parabola) This has 3 different polynomials for x, and will touch or cross the x axis 3 times
Roots of a polynomial are locations where y=0. Rule of signs can tell us how many roots exist, and how many have a positive or negative value of x.
The highest power is the number of total roots.
Count the number of times the sign of coefficient of terms changes sign (ignoring any zero coefficients) as you go down the list of coefficients.
If there is no constant term (power of 0), pull out a factor of x, until there is a constant. For example: x³+2x²+5x gets changed to x(x²+2x+5), then use x²+2x+5 in the steps below. (Total number of roots is from original equation: 3 in this case.)
Example: -3x⁴ + 4x² + x − 2 Positive Roots: This has two changes in sign, so a maximum of two positive roots Negative Roots: Substitute -x => only odd powers will change sign. Count the changes in sign again. In this example, only the x changes: -3(-x)⁴ + 4(-x)² + -x − 2 = -3x⁴ + 4x² -x − 2 still two changes => two negative roots (or 0 if there's complex roots) The number of positive or negative roots is reduced by 2 for each pair of complex roots, if they exist. That means, in this case, there could be: * 4 roots: 2 positive roots, 2 negative roots * 4 roots: 0 positive roots, 2 negative roots, complex pair * 4 roots: 2 positive roots, 0 negative roots, complex pair * 4 roots: 0 positive roots, 0 negative roots, 2 complex pairs
What are all the possible rational roots of the following function: 6x⁴−x³−4x²−x−2
Unpacking this: first need synthetic division for polynomials: https://courses.lumenlearning.com/waymakercollegealgebra/chapter/synthetic-division/
Remainder theorem: If a polynomial f(x) is divided by x – k, then the remainder is the value f(k).
⇒ f(x)/(x-k) = f(k), or f(x)/f(k) = x-k
Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial, but first we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial.
The Rational Zero Theorem states that if the polynomial f(x)=anx^n+an−1x^n−1+…+a1x+a0 (coefficients are a0, a1, .. an) has integer coefficients, then every rational zero of f(x) has the form pq where p is a factor of the constant term a0 and q is a factor of the leading coefficient an. =⇒ So dividing all the p’s by all the q’s give all the possible rational zeros.
When the leading coefficient is 1, the possible rational zeros are the factors of the constant term.