Topic VII: Polynomial Functions 7.3 Solving Polynomial Equations.

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Presentation transcript:

Topic VII: Polynomial Functions 7.3 Solving Polynomial Equations

FACTORING AND ROOTS CUBIC FACTORING a³ + b³ = (a + b)(a² - ab + b²) a³ - b³ = (a - b)(a² + ab + b²) Difference of Cubes Sum of Cubes Question: if we are solving for x, how many possible answers can we expect? 3 because it is a cubic!

CUBIC FACTORING EX- factor and solve 8x³ - 27 = 0 (2x)³ - (3) 3 = (2x - 3)((2x)² + (2x)3 + 3²) (2x - 3)(4x² + 6x + 9)=0 Quadratic FormulaX= 3/2 a³ - b³ = (a - b)(a² + ab + b²)

CUBIC FACTORING EX- factor and solve x³ = 0 a³ + b³ = (a + b)(a² - ab + b²) x³ = (x + 7)(x² - 7x + 7²) (x + 7)(x² - 7x + 49)=0 Quadratic FormulaX= -7

Let’s try one Factor a)x 3 -8b)x

Let’s try one Factor a)x 3 -8b)x

Let’s Try One 81x =0 Hint: IS there a GCF???

Let’s Try One 81x =0

Factor by Using a Quadratic Form Ex: x 4 -2x 2 -8 Since this equation has the form of a quadratic expression, we can factor it like one. We will make temporary substitutions for the variables = (x 2 ) 2 – 2(x 2 ) – 8 Substitute a in for x 2 = a 2 – 2a – 8 This is something that we can factor (a-4)(a+2) Now, substitute x 2 back in for a (x 2 -4)(x 2 +2) (x 2 -4) can factor, so we rewrite it as (x-2)(x+2) So, x 4 -2x 2 -8 will factor to (x-2)(x+2)(x 2 +2)

Let’s Try One Factor x 4 +7x 2 +6

Let’s Try One Factor x 4 +7x 2 +6

Let’s Try One Where we SOLVE a Higher Degree Polynomial x 4 -x 2 = 12

Let’s Try One Where we SOLVE a Higher Degree Polynomial x 4 -x 2 = 12 Solutions

Ex : Solve using the Rational Root Theorem:

–1 1 7–9 1 –1 –2 2 9 –9 –18 Find all real zeros of the function. 4. Possible rational zeros:  1,  18  2,  9,  3,  6, – (x + 1) (x + 1)(x 2 + 9)(x – 2) = 0 x 3 – 2x 2 + 9x – 18 ( ) = 0( ) x2x2 (x – 2) +9(x – 2) (x 2 + 9) (x – 2) (x + 1) x =

Write a polynomial function of least degree with a leading coefficient of 1 whose zeroes are: 2, 3, -4 f(x) = (x – 2)(x – 3)(x + 4) f(x) =(x 2 – 5x + 6)(x + 4) f(x) = x 3 - x 2 – 14x + 24

Write a polynomial function of least degree with a leading coefficient of 1 whose solutions are: 7, i f(x) = (x – 7)(x – i)(x + i) f(x) = (x – 7)(x 2 + 1) f(x) = x 3 – 7 x 2 + x – 7

Now we are ready 2 work