PERIOD Lesson 5-6 PDF Pass Chapter 5 37 Glencoe Algebra 2 1 f(x) = x2 + 6x + 5 Skills Practice The Remainder and Factor Theorems 5-6
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The Remainder and Factor Theorems Goals p Divide polynomials and relate the result to the 5 3 2 2 6 1 1 3 9 Your Notes REMAINDER THEOREM
Alg2_6_5.pdf
So, -3x + 15x² + 108x – 540 = (x - 6)(-3x² – 3x +90) Factoring the quadratic expression yields –3x2 + 15x + 108x – 540 = -3(x – 6)(x + 6)(x – 5
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This means that we no longer need to write the quotient polynomial down, nor the x in the divisor, to determine our answer -2 x3+4x2- 5x -14 2x2 12x 14 x3
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The Remainder Theorem states that f(2) will be equal to 65 Try it f(2)= (2)3+2(2)2-6(2)-9 f+8-12-9 = 16-12-a = 4-9 = -5 - it works Factor Theorem:
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and factor theorems to find factors of polynomials expression there is a remainder 5)12 3)(2()35 4 3( 5 2 3 4 2 5 2 6
A26remainder.pdf
For Problems 12 – 16, use synthetic division and the Remainder Theorem to find the indicated function value 12 )3(;6 5 7 )( 2 3
mth103fa13_chapter5.pdf
3 ? 17a 2 + 2a ? 7 at a = 3 6) f (x) = x 5 ? 47x 3 ? 16x 2 + 8x + 52 at x = 7 State if the given binomial is a factor of the given polynomial
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remainder is 0, then the divisor is a factor of the dividend Step 5 Continue as in Steps 1-4, but in Step 2 divide the leading term of the expres-
synthetic_division.pdf
5-6 Skills Practice The Remainder and Factor Theorems L Use synthetic substitution to find f(2) and f(-1) for each function 1,1x) = x2 + 6x + 5
Alg%202%20Chapter%205%20Part%202%20(5-5%20through%205-8)%20Study%20Guide%20&%20Key.pdf