1.9.0 Long division of polynomials - MHF4U Hyde 1.9.0 Long Division of Polynomials Review of long division and terms 2875 8=359R3 28i=3593=359 3 quuoent

# 1.9.0 Long division of polynomials - MHF4U Hyde 1.9.0 Long...

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This preview shows page 1 - 2 out of 2 pages. Subscribe to view the full document. 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°&gt;\// 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&lt;H+OA&lt;3 *B’Z’H, l ,«Tqr’xv‘d-f 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&amp;3 L 2 5 - Hog/H24 x—intercepts are 2,——,and— 3 2 — (~ ”A” +22% V . . y-Intercept IS 20 ...
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