140 likes | 274 Views
Long Division. Just like in the 3 rd grade 6x 3 + 19x 2 + 16x – 4 divided by x - 2 x - 2 6x 3 + 19x 2 + 16x - 4. U-Try. (4x 3 -7x 2 – 11x + 5) divided by (4x + 5). Synthetic division. to divide ax 3 + bx 2 + cx +d by (x – k). k. a b c d. ka. k(b-ka).
E N D
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