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Exercise 1.2 · Q78

Q.Find the value of 2x3−11x2+44x+272x^3-11x^2+44x+27, if x=253−4ix = \dfrac{25}{3-4i}.

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Simplify x=dfrac253−4ix=\\dfrac{25}{3-4i} by rationalising: x=dfrac25(3+4i)(3−4i)(3+4i)=dfrac25(3+4i)9+16=dfrac25(3+4i)25=3+4ix=\\dfrac{25(3+4i)}{(3-4i)(3+4i)}=\\dfrac{25(3+4i)}{9+16}=\\dfrac{25(3+4i)}{25}=3+4i. Now substitute x=3+4ix=3+4i into 2x3−11x2+44x+272x^3-11x^2+44x+27. x2=(3+4i)2=9+24i+16i2=9+24i−16=−7+24ix^2=(3+4i)^2=9+24i+16i^2=9+24i-16=-7+24i. x3=x2x=(−7+24i)(3+4i)=−21−28i+72i+96i2=−21+44i−96=−117+44ix^3=x^2x=(-7+24i)(3+4i)=-21-28i+72i+96i^2=-21+44i-96=-117+44i. So $2x^ …

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