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Unformatted text preview: MHF4U Hyde 1.9.0 Long Division of Polynomials
Review of long division and terms. 2875+8=359R3 28i=3593=359+3 quuoﬁent+w
8 8 8 divisor divisor
359R3
8 2875 or 2875 = 8(359)+ 3 dividend = quotient(divis0r)+ remainder 8 goes into 28 3 times. 3 times 8 is 24. 28 minus 24 is 4. Bring down the
7, and so on. We can use the same strategies to divide polynomials. Ex1. Let’s consider %. We would normally factor first. Let’s try long
division.
rz4’1
x + 3 x2 + 5x + 6
_. ,/ «z+‘r°>\//
ﬁJré x goes into x2 x times. Now we distribute x through
Jz/xra x+3 , giving us x2 +3x, and subtract from x2 +5x,
J leaving 3x+6. x goes into 3x 3 times. Now we
0 distribute 3 through x+3, giving us 3x+ 6, and subtract
from 3x+ 6, leaving a remainder of zero. The
remainder of zero means that x + 3 is a factor of
x2 + 5x + 6.
. . 3x3—5x2+6x—7
Ex2. Simplify —— x2 +4x+1 3x3 ~5x2 +6x—7 71x+10
—§‘_—=3x—17+—2——
x +4x+1 x +4x+1 or
3x3 ~5x2 +6x—7 = (x2 +4x+1)(3x—17)+71x+10 MHF4U Hyde x4—4x2+x+4
x2—3 Introduce visible coefficients of zero. Ex3. Simplify {1 4_ 2
x2+0x—3ix4+0x3—4x2+x+4 wi=x2—l+ J:+1
x —3 x —3 __(9<H+OA<3 *B’Z’H, l ,«Tqr’xv‘df 6(4)” or x4 —4x2 +x+4=(x2 —1x2 —3)+x+1 Ex4. x  2 is a factor of f(x) = 6x3 — 23x2 + 12x + 20. Find the other factors,
determine all intercepts, and sketch f . 6x2 —11x—10 x—2j6x3 —23x2 +12x+20 , so f(x)=(x—2)(6x2 —11x—10)=(x—2)(3x+2)(2x—5) J5“ 4&3 L 2 5  Hog/H24 x—intercepts are 2,——,and— 3 2
— (~ ”A” +22% V . .
yIntercept IS 20 ...
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 Spring '15
 Division, 3 L, 22%, MHF4U Hyde

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