(x^3+5x+2x-8)÷x+2
divide by long division method
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hayy mate here your answer ✔️ ✔️
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As the coefficient of x3 in x3−5x2+2x+8=0 is 1,
One of the roots must be factor of 81=8 i.e. it can be ±1.±2,±4or±8.
As sum of coefficients (1−5+2+8=6≠0) is not zero, it is evident that 1 is not the root of x3−5x2+2x+8=0.
But −1 is a root as −1−5−2+8=0 and hence as per factor theorem (x+1) is a factor of x3−5x2+2x+8=0
Dividing x3−5x2+2x+8 by (x+1) we get
x3−5x2+2x+8=x2(x+1)−6x(x+1)+8(x+1)=(x+1)(x2−6x+8)
now we can further factorize x2−6x+8 by splitting middle term
x2−6x+8=x2−4x−2x+8=x(x−4)−2(x−4)=(x−2)(x−4)
Hence, roots of x3−5x2+2x+8=0 are {−1,2,4}.
____________________________
❤️⭐I hope you mark as brainlist answer⭐❤️✨✨✨
ANY MISTAKE THEN COMMENTS ME I EDIT MY ANSWER ✅✅
_____________________________
As the coefficient of x3 in x3−5x2+2x+8=0 is 1,
One of the roots must be factor of 81=8 i.e. it can be ±1.±2,±4or±8.
As sum of coefficients (1−5+2+8=6≠0) is not zero, it is evident that 1 is not the root of x3−5x2+2x+8=0.
But −1 is a root as −1−5−2+8=0 and hence as per factor theorem (x+1) is a factor of x3−5x2+2x+8=0
Dividing x3−5x2+2x+8 by (x+1) we get
x3−5x2+2x+8=x2(x+1)−6x(x+1)+8(x+1)=(x+1)(x2−6x+8)
now we can further factorize x2−6x+8 by splitting middle term
x2−6x+8=x2−4x−2x+8=x(x−4)−2(x−4)=(x−2)(x−4)
Hence, roots of x3−5x2+2x+8=0 are {−1,2,4}.
____________________________
❤️⭐I hope you mark as brainlist answer⭐❤️✨✨✨
ANY MISTAKE THEN COMMENTS ME I EDIT MY ANSWER ✅✅
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