Determine the zeros of each fynction 1. f(x)=x³+1 2. f(x) =x⁴-1 3. f(x)=x³-5x²+6x 4.f(x)=x⁵-10x³+9x 5.f(x)=x³-x²-5x+5 6.f(x) =x³-3x²-9x-5 7. f(x)=x⁵-3x⁴+x³+x²+4
Answers
Step-by-step explanation:
Solution

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Given equation, x5−5x4+9x3−9x2+5x−1=0
Consider f(x) = x^5-5x^4+9x^3-9x^2+5x-1
Notice that this is a reciprocal equation of odd degree which has the opposite signs of the first and last term.
∴(x−1) is one factor of the given equation and the quotient is another reciprocal function which has same signs of the first and last term.
∴f(x)=(x−1)(Ax4+Bx3+Cx2+Bx+A)
Comparing the coefficient, we have A=1,B=−4,C=5
⟹f(x)=(x−1)(x4−4x3+5x2−4x+1)
Consider g(x)=x4−5x3−22x2−5x+1=(x4+1)−4(x3+x)+5x2
We need to find the roots of g(x)=0
⟹(x2+x−2)–4(x+x−1)+5=0[dividing byx2]
Substitute x+x−1=y in the above equation
⟹(y2−
Answer:
(6xy) × (-3x²y³)
= {6 × (-3)} × {xy × x²y³}
= -18x1+2 y1+3
= -18x³y⁴.
(ii) 7ab², -4a²b and -5abc
Solution:
(7ab²) × (-4a²b) × (-5abc)
= {7 × (-4) × (-5)} × {ab² × a²b × abc}
= 140 a1+2+1 b2+1+1 c
= 140a⁴b⁴c.
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Step-by-step explanation:
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