Chapter 8Chapter 8

Variation and Variation and Polynomial Polynomial EquationsEquations

Section 8-1Section 8-1

Direct Variation Direct Variation and Proportionand Proportion

Direct VariationDirect Variation

A linear function A linear function defined by an defined by an equation of the form equation of the form y = mxy = mx

y varies directly as xy varies directly as x

Constant Constant VariationVariation

The constant The constant mm is the is the constant variationconstant variation

Example 1Example 1 The stretch is a loaded The stretch is a loaded spring varies directly as spring varies directly as the load it supports. A the load it supports. A load of 8 kg stretches a load of 8 kg stretches a certain spring 9.6 cm. certain spring 9.6 cm.

Find the constant of Find the constant of variation (m) and the variation (m) and the equation of direct equation of direct variation.variation.

m = 1.2m = 1.2 y = 1.2xy = 1.2x What load would stretch What load would stretch the spring 6 cm?the spring 6 cm?

5 kg5 kg

ProportionProportion An equality of ratiosAn equality of ratios

yy11 = y = y22

xx11 x x22

Directly Directly ProportionalProportional

In a direct variation, y In a direct variation, y is said to be directly is said to be directly proportional to xproportional to x

Constant of Constant of ProportionalityProportionality

m is the constant of m is the constant of proportionalityproportionality

Means and Means and ExtremesExtremes

meansmeansyy11:x:x11 = y = y22:x:x22

extremes

Solving a Solving a ProportionProportion

The product of the The product of the extremes equals the extremes equals the product of the meansproduct of the means

yy11xx22 = y = y22xx11

To get this product, To get this product, cross multiplycross multiply

Example 2Example 2

If y varies directly as If y varies directly as x, and y = 15 when x, and y = 15 when x=24, find x when y = x=24, find x when y = 25.25.

x = 40x = 40

Example 3Example 3 The electrical resistance The electrical resistance in ohms of a wire varies in ohms of a wire varies directly as its length. If a directly as its length. If a wire 110 cm long has a wire 110 cm long has a resistance of 7.5 ohms, resistance of 7.5 ohms, what length wire will have what length wire will have a resistance of 12 ohms?a resistance of 12 ohms?

Section 8-2Section 8-2

Inverse and Inverse and Joint VariationJoint Variation

Inverse VariationInverse VariationA function defined by A function defined by an equation of the form an equation of the form xy = kxy = k or or y = k/xy = k/x

y varies inversely as x, y varies inversely as x, or y is inversely or y is inversely proportional to xproportional to x

Example 1Example 1

If y is inversely If y is inversely proportional to x, and proportional to x, and y = 6 when x = 5, find y = 6 when x = 5, find x when y = 12.x when y = 12.

x = 2.5x = 2.5

Joint VariationJoint Variation

When a quantity When a quantity varies directly as the varies directly as the product of two or product of two or more other quantitiesmore other quantities

Also called Also called jointly jointly proportionalproportional

Example 2Example 2 If z varies jointly as x If z varies jointly as x and the square root and the square root of y, and z = 6 when x of y, and z = 6 when x = 3 and y = 16, find z = 3 and y = 16, find z when x = 7 and y = 4.when x = 7 and y = 4.

z = 7z = 7

Example 3Example 3 The time required to travel a The time required to travel a given distance is inversely given distance is inversely proportional to the speed of proportional to the speed of travel. If a trip can be made in travel. If a trip can be made in 3.6 h at a speed of 70 km/h, 3.6 h at a speed of 70 km/h, how long will it take to make how long will it take to make the same trip at 90 km/h?the same trip at 90 km/h?

Section 8-3Section 8-3

Dividing Dividing PolynomialsPolynomials

Long DivisionLong DivisionUse the long division Use the long division process for polynomialsprocess for polynomials

Remember:Remember:

873 ÷ 14 = ?873 ÷ 14 = ?62 5/1462 5/14

Example 1Example 1

DivideDivide

xx33 – 5x – 5x22 + 4x – 2 + 4x – 2

x – 2x – 2xx22 – 3x – 2 + -6/x-2 – 3x – 2 + -6/x-2

CheckCheck

To check use the To check use the algorithm:algorithm:

Dividend = (quotient)Dividend = (quotient)(divisor) + remainder(divisor) + remainder

Section 8-4Section 8-4

Synthetic Synthetic DivisionDivision

Synthetic DivisionSynthetic Division

An efficient way to An efficient way to divide a polynomial divide a polynomial by a binomial of the by a binomial of the form form x – cx – c

Reminder:Reminder:

The divisor must be in The divisor must be in the form the form x – cx – c

If it is not given in If it is not given in that form, put it into that form, put it into that formthat form

Example 1Example 1

Divide:Divide:

xx44 – 2x – 2x33 + 13x – 6 + 13x – 6

x + 2x + 2 xx33 – 4x – 4x22 + 8x - 3 + 8x - 3

Section 8-5Section 8-5

The Remainder The Remainder and Factor and Factor TheoremsTheorems

Remainder Remainder TheoremTheorem

Let P(x) be a polynomial Let P(x) be a polynomial of positive degree of positive degree n.n. Then for any number Then for any number cc, , P(x) = Q(x)(x – c) + P(c) P(x) = Q(x)(x – c) + P(c) where Q(x) is a where Q(x) is a polynomial of degree n-1.polynomial of degree n-1.

Remainder Remainder TheoremTheorem

You can use synthetic You can use synthetic division as “synthetic division as “synthetic substitution” in order substitution” in order to evaluate any to evaluate any polynomialpolynomial

Synthetic Synthetic SubstitutionSubstitution

Evaluate at P(-4)Evaluate at P(-4)

P(x) = xP(x) = x44 – 14x – 14x22 + 5x – 3 + 5x – 3 Use synthetic division Use synthetic division to find the remainder to find the remainder when c = -4 when c = -4

Factor TheoremFactor Theorem

The polynomial P(x) The polynomial P(x) has has x – rx – r as a factor if as a factor if and only if and only if rr is a root is a root of the equation P(x) = of the equation P(x) = 00

ExampleExampleDetermine whether Determine whether x + 1 is a factor of x + 1 is a factor of P(x) = xP(x) = x1212 – 3x – 3x88 – 4x – 2 – 4x – 2 If P(-1) = 0, then x + If P(-1) = 0, then x + 1 is a factor1 is a factor

ExampleExampleFind a polynomial equation Find a polynomial equation with integral coefficients that with integral coefficients that has 1, -2 and 3/2 as rootshas 1, -2 and 3/2 as roots

The polynomial must have The polynomial must have factors (x – 1), (x – (-2)) and (x factors (x – 1), (x – (-2)) and (x – 3/2).– 3/2).

Depressed Depressed EquationEquation

Solve xSolve x33 + x + 10 = 0, + x + 10 = 0, given that -2 is a rootgiven that -2 is a root

To find the solution, To find the solution, divide the polynomial divide the polynomial by x – (-2)by x – (-2)