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Dive into the world of polynomial division with this comprehensive guide focusing on long division, synthetic division, and the Remainder Theorem. Learn how to divide polynomials such as (6x^3 + 19x^2 + 16x - 4) by (x - 2) and explore practical examples like (4x^3 - 7x^2 - 11x + 5) divided by (4x + 5). Understand how to find remainders efficiently and test for roots of polynomials. Ideal for students seeking to reinforce their skills from the 3rd grade onward, this guide simplifies complex instructions into easy-to-follow steps.
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Long Division • Just like in the 3rd grade 6x3 + 19x2 + 16x – 4 divided by x - 2 x - 2 6x3 + 19x2 + 16x - 4
U-Try (4x3 -7x2 – 11x + 5) divided by (4x + 5)
Synthetic division to divide ax3 + bx2 + cx +d by (x – k) k a b c d ka k(b-ka) a b-ka remainder Coefficients of Quotient
Lets try one x4 – 10x2 – 2x + 4 divided by ( x + 3) ( x + 3) k = -3 x4 + 0x3 – 10x2 – 2x + 4 1 0 –10 – 2 4 -3 -3 9 3 -3 1 -3 -1 1 1
x4 + 0x3 – 10x2 – 2x + 4 1 0 –10 – 2 4 -3 -3 9 3 -3 remainder 1 -3 -1 1 1 __ __x3 + __x2 + __x + __ + (x + 3)
You try • 3x3 -17x2 + 15x -25 divided by (x - 5)
Pretty Cool Remainder Theorem Or the PCRT • If a polynomial is divided by (x - k) then the remainder will be f(k)
Let’s try one • Find the remainder of the problem • 9x3 – 16x – 18x2 + 32 divided by (x – 2) f(x) = 9x3 – 16x – 18x2 + 32 f(2) = 9(2)3 – 16(2) – 18(2)2 + 32 f(2) = 9(8) – 16(2) – 18(4) + 32
Is it a root? • If you try synthetic division and there is no remainder, that means k is a solution to f(x) = 0 or … f(k) = 0 A great way to test for roots of higher degree polynomials
The most confusing instructions for any homework problem I’ve ever seen! What they want Take f(x) and divide it by (x - k) Then write (x - k) (quotient) + remainder (x-k) Page 233 problems 39 -46
Divide then divide again to factor 3rd degree polynomials given 2 factors
Lets try a few problems • Page 235 problems • 7 - 15 • 21 - 27 • 51 - 65