Second-Degree Polynomials

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Tables 22-1 through 22-5 show how we can convert the above equations to polynomial standard form Note the last step in Table 22-5 To be true to form, the polynomial should show the terms by descending powers of the variable The term containing x 2 should come first, then the term containing x, and finally the constant

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Table 22-1

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Statements x 2 = 2x + 3 x 2 2x = 3 x 2 2x 3 = 0

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Conversion of x 2 = 2x + 3 to polynomial standard form

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Reasons This is the equation we are given Subtract 2x from each side Subtract 3 from each side

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Table 22-2

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Statements x = 4x 2 7 4x 2 + x = 7 4x 2 + x + 7 = 0

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Conversion of x = 4x 2 7 to polynomial standard form

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Reasons This is the equation we are given Subtract 4x 2 from each side Add 7 to each side

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Table 22-3 Conversion of x 2 + 4x = 7 + x to polynomial standard form

Statements x 2 + 4x = 7 + x x 2 + 4x 7 = x x 2 + 3x 7 = 0 Reasons This is the equation we are given Subtract 7 from each side Subtract x from each side

Table 22-4 Conversion of x 2 = 8x 2 22 to polynomial standard form

Statements x 2 = 8x 2 22 8x 2 + x 2 = 22 8x 2 + x + 20 = 0 Reasons This is the equation we are given Add 8x 2 to each side Add 22 to each side

Table 22-5

Statements 3 + x = 2x 2 2x 2 + 3 + x = 0 2x 2 + x + 3 = 0

Conversion of 3 + x = 2x 2 to polynomial standard form

Reasons This is the equation we are given Subtract 2x 2 from each side Commutative law for addition

366 Quadratic Equations with Real Roots

Are you confused

Suppose you see an equation in one variable You think it s a quadratic, but you aren t sure Here s an example: 3x 2 + x 8 = 4x 2 + 7x + 4 If you can convert this equation to polynomial standard form, you can be certain that it s a quadratic If you can t convert it, then two things are possible: you didn t try hard enough, or it isn t a quadratic If it isn t a quadratic and you want to prove that it isn t, you must morph the equation into a form that s obviously not convertible into polynomial standard form That can be tricky The next challenge will show you an example As things work out, you can convert the above equation to polynomial standard form If you subtract 4x 2 from each side, then subtract 7x from each side, and finally subtract 4 from each side, you get x 2 6x 12 = 0

Here s a challenge!

Show that the following is the equivalent of a first-degree equation, not a quadratic: 7x 2/2 + 7x 5 = 23x 2/( 10) + 6x 2/5 + 3x

Solution

It takes a little intuition to solve this, but nothing beyond the mathematical skill we ve acquired by now! Let s look closely at the right side of this equation It contains two terms with x 2 These terms are: 23x 2/( 10) and 6x 2/5 We can add these two terms to get a single term in x 2 If we multiply both the numerator and denominator of the first of the above expressions by 1, and if we multiply both the numerator and denominator of the second expression by 2, we get 23x 2/10 and 12x 2/10 We now have a common denominator, so we can find the sum 23x 2/10 + 12x 2/10 = 35x 2/10

Binomial Factor Form

which reduces to 7x 2/2 This allows us to rewrite the right side of the original equation to obtain 7x 2/2 + 7x 5 = 7x 2/2 + 3x Subtracting 7x 2/2 from each side, we get 7x 5 = 3x

When we subtract 3x from each side now, we get an equation in the standard single-variable, first-degree form: 4x 5 = 0

Binomial Factor Form

There s another way to express a quadratic equation: as a product of two binomials that is equal to 0 In some ways, this form is simpler than the polynomial standard form As you ll soon see, the binomial factor form of a quadratic tells you the roots directly

Binomials in quadratics When a left side of a quadratic is expressed as a product of binomials, both of the binomials must be in a specific form The first term in each binomial is a multiple of the variable The multiplicand in that term is the coefficient of the variable The second term is a constant, sometimes called the stand-alone constant Here are some examples: