Polynomials with complex roots .au



PolynomialsPolynomials with real coefficients and complex rootsFactorise and determine the roots to the polynomial Px=x3-3x2+6x-4P1=1-3+6-4=0x=1 is a zero and x-1 is a factor of the polynomial.Divide Px by (x-1) to obtain:Px=x3-3x2+6x-4=(x-1)(x2-2x+4)Solve x2-2x+4=0x=-b±b2-4ac2aSubstitute a=1, b=-2 and c=4x=-(-2)±(-2)2-4×1×42×1=2±4-162=2±-122=2±-1×4×32=2±23i2=1±3iRoots of the polynomial: x=1, 1±3iFactorise and determine the roots to the polynomial Px=x5-3x3-4xPx=x5-3x3-4x=xx4-3x2-4=xx2-4)(x2+1Solve xx2-4)(x2+1=0x=0,x2-4=0,x=±2x2+1=0and x=±iRoots of the polynomial: x=0, ±2, ±iFactorise and determine the roots to the polynomial Px=x3+2x2+2x+4Method 1: Try to find an integer zero/root by considering the polynomial and the product of the roots-da=-41=-4Factors of -4: ±1, ±2, ±4P(x)≠0 if x>0 as all coefficients are positive.Consider -1, -2, -4.P-1=-13+2-12+2-1+4=3P-2=-23+2-22+2-2+4=0P-4=-43+2-42+2-4+4=-36x=-2 is a zero and x+2 is a factor of the polynomial.Divide Px by (x+2) to obtain:Px=x3+2x2+2x+4=(x+2)(x2+2)Solve (x+2)(x2+2)=0x+2=0, and x2+2=0x=-2 and x=±√2iRoots of the polynomial: x=0, ±√2iMethod 2: Factorise by grouping in pairsPx=x3+2x2+2x+4=x2x+2+2x+2=(x+2)(x2+2)Solve as above.Factorise and determine the roots to the polynomial Px=2x4+2x3-11x2-4x+20-da=202=10Factors of 10: ±1, ±2, ±5,±10Try to find an integer zero/root by inspection:P1=214+213-1112-41+20=9P-1=2-14+2-13-11-12-4-1+20=-19P2=224+223-1122-42+20=16P-2=2-24+2-23-11-22-4-2+20=0x=-2 is a zero and x+2 is a factor of the polynomial.Divide Px by (x+2) to obtain:Px=2x4+2x3-11x2-4x+20=(x+2)(2x3-2x2-7x+10)Try to find an integer zero/root by inspection for 2x3-2x2-7x+10P1=213-212-71+10 =3P-1=2-13-2-12-7-1+10 =13P2=223-222-72+10 =4P-2=2-23-2-22-7-2+10 =0x=-2 is a zero and x+2 is a factor of the polynomial.Check x+2 is a multiple rootP'x=8x3+6x2-22x-4P'-2=8-23+6-22-22-2-4=0P''x=24x2+12x-22P''-2=24-22+12-2+22=94∴x+2 is a double root.Divide Px by x+22 to obtain:Px=2x4+2x3-11x2-4x+20=(x+2)2(2x2-6x+5)Solve 2x2-6x+5=0x=-b±b2-4ac2aSubstitute a=2, b=-6 and c=5x=-(-6)±(-6)2-4×2×52×2=6±36-404=6±-44=6±-1×44=6±2i4=32±i2Roots of the polynomial: x=-2, 32±i2Given 2-3i is a zero of Px=x4-7x3+27x2-47x+26 find all roots of p(x).Method 1: Using division of the polynomialThe polynomial has real coefficients and therefore the complex roots will always occur in conjugate pairs.∴x=2+3i is also a root/zero of the polynomial.x-(2-3i) and x-(2+3i) are factors of the polynomial.Px=x4-7x3+27x2-47x+26=x-2-3ix-2+3iQ(x)=(x2-4x+13)Q(x)Divide Px by (x2-4x+13) to find Q(x):Px=x4-7x3+27x2-47x+26=(x2-4x+13)(x2-3x+2)Factorise x2-3x+2=(x2-4x+13)(x-2)(x-1)Roots of the polynomial: x=1, 2, 2-3i, 2+3iMethod 2: Using the sum and products of rootsThe polynomial has real coefficients and therefore the complex roots will always occur in conjugate pairs.∴x=2+3i is also a root/zero of the polynomial.Let the other two roots be α and βSum of the roots =-ba: α+β+2+3i+2-3i= 7α+β+4=7α+β=3β=3-αProduct of the roots =da: αβ2+3i2-3i=2613αβ=26αβ= 2α3-α=23α-α2=20=α2-3α+20=(α-2)(α-1)α=2 or 1α=2β=3-2=1α=1β=3-1=2Roots of the polynomial: x=1, 2, 2-3i, 2+3i ................
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