# Polynomials Expressions like 3x 4 + 2x 3 – 6x 2 + 11 and m 6 – 4m 2 +3 are called polynomials. (5x – 2)(2x+3) is also a polynomial as it can be written.

## Presentation on theme: "Polynomials Expressions like 3x 4 + 2x 3 – 6x 2 + 11 and m 6 – 4m 2 +3 are called polynomials. (5x – 2)(2x+3) is also a polynomial as it can be written."— Presentation transcript:

Polynomials Expressions like 3x 4 + 2x 3 – 6x 2 + 11 and m 6 – 4m 2 +3 are called polynomials. (5x – 2)(2x+3) is also a polynomial as it can be written 10x 2 + 11x - 6. The degree of the polynomial is the value of the highest power. 3x 4 + 2x 3 – 6x 2 + 11 is a polynomial of degree 4. m 6 – 4m 2 + 3 is a polynomial of degree 6. In the polynomial 3x 4 + 2x 3 – 6x 2 + 11, the coefficient of x 4 is 3 and the coefficient of x 2 is -6.

A root of a polynomial function, f(x), is a value of x for which f(x) = 0

Hence the roots are 2 and –2.

Division by (x – a) Dividing a polynomial by (x – a) allows us to factorise the polynomial. You already know how to factorise a quadratic, but how do we factorise a polynomial of degree 3 or above? We can divide polynomials using the same method as simple division. 4 32 30 3 24 6 Conversely, Divisor Quotient Remainder

Divisor Quotient Remainder Note: If the remainder was zero, (x – 2) would be a factor. This is a long winded but effective method. There is another.

Let us look at the coefficients. 2 5 2 8 -2 5 10 12 24 32 64 62 quotient divisor remainder This method is called synthetic division. If the remainder is zero, then the divisor is a factor. Let us now look at the theory.

We divided this polynomial by (x – 2). 62 If we let the coefficient be Q(x), then Remainder Theorem If a polynomial f (x) is divided by (x – h) the remainder is f (h). We can use the remainder theorem to factorise polynomials.

Remainder Theorem If a polynomial f (x) is divided by (x – h) the remainder is f (h). 42 -9 5 -3 -4 2 8 -4 1 4 1 4 0 Since the remainder is zero, (x – 4) is a factor.

To find the roots we need to consider the factors of -15. 3 2 5 -28 -15 2 6 11 33 5 15 0

To find the roots we need to consider the factors of -18. 3 1 0 -7 0 -18 1 3 3 9 2 6 6 18 0 -3 1 3 2 6 1 -3 0 0 2 -6 0

Finding a polynomial’s coefficients We can use the factor theorem to find unknown coefficients in a polynomial. Since we know (x + 3) is a factor, the remainder must be zero. -3 2 6 p 4 -15 2 -6 0 0 p -3p 4 - 3p 9p - 12 9p - 27

2 1 a -1 b -8 1 2 2 + a 4 + 2a 3 + 2a 6 + 4a 6 + 4a + b 12 + 8a + 2b 4 + 8a + 2b = 0 -4 1 a -1 b -8 1 -4 a - 4 16 - 4a 15 - 4a 16a - 60 16a + b - 60 240 - 64a - 4b 232 - 64a - 4b = 0

Solving polynomial equations 3 1 -4 1 6 1 3 -3 -2 -6 0 If we sketch the curve of f (x), we see that the roots are where f (x) crosses the X axis.

Functions from Graphs f (x) x d a b c The equation of a polynomial may be established from its graph.

1. From the graph, find an expression for f (x). f (x) x -12 -3 2 Substituting (0, -12)

2. From the graph, find an expression for f (x). f (x) x 30 -2 1 5 Substituting (0, 30)

Curve sketching The factor theorem can be used when sketching the graphs of polynomials. The Y axis intercept is (0, 12) Y axis Intercept

We will use synthetic division to find the roots of the function. This will tell us where the graph crosses the X axis. 2 1 -1 -8 12 1 2 1 2 -6 -12 0 X axis Intercept

Stationary Points

Nature of Stationary Points Slope - - + 0 + - - 0 + + +

Approximate Roots When the roots of f (x) = 0 are not rational, we can find approximate values by an iterative process. We know a root exists if f (x) changes sign between two values. f (x) x a b f (x) x a b A root exists between a and b.

Hence the graph crosses the x - axis between 1 and 2. 1 2 2 -5 1 and 2 1.5 -1.625 1 and 1.5 1.4 -0.896 1 and 1.4 1.3 -0.163 1 and 1.3 1.2 0.568 1.2 and 1.3 1.25 0.203 1.25 and 1.3 1.26 0.130 1.26 and 1.3 1.27 0.057 1.27 and 1.3 1.28 -0.016 1.27 and 1.28 1.275 0.020 1.275 and 1.28 Hence the root is 1.28 to 2 d.p.

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