Question

find the zeroes of polynomial px =x³-5x²-24x if it is given that product of its two zeroes is -8​

Answer 1

Answer:

X^3-6x^2+3x-1 has real roots p, q, and r. What is p^2q+q^2r+r^2p?

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x3−6x2+3x−1 has real roots p, q, and r. What is p2q+q2r+r2p?

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x3−6x2+3x−1(1)

p,q,r being roots

p+q+r=6

pq+qr+rp=3

pqr=1

let A=p2q+q2r+r2p

A′=pq2+qr2+rp2

A+A′=pq(p+q)+qr(q+r)+rp(r+p)

=pq(6−r)+qr(6−p)+rp(6−q)

=6(pq+qr+rp)−3pqr

=6×3−3=15

AA′=(p2q+q2r+r2p)(pq2+qr2+rp2)

=3+(p3+q3+r3)+1p3+1q3+1r3 After some simplification

p3+q3+r3=(p+q+r)3−3(pq+qr+rp)(p+q+r)+3pqr

=63−3×3×6+3=165

1p3+1q3+1r3=−24 likewise

Hence AA′=3+165–24=144

Hence A and A’ are roots of y2–15y+144=0

A=15+i351−−−√2

or

A=15−i351−−−√2

So the required value is not real. as claimed