Question

Verify that 1, –1 and –3 are the zeros of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeros and the coefficients.

Answer 1

Cubic:

$ax^{3} +bx^{2} +cx+d$

Where, a,b,c,d are coefficients

Let p,q and r are the zeros of the polynomial.

Then, p+q+r= -b/a

And, pqr= -d/a

pq+qr+rp = c/a

i)$x^{3} +3x^{2} -x-3$

$*1^{3} +3*1^{2} -1*1-3$

1+3-1-3=0

$x^{3} +3x^{2} -x-3$

$*(-1)^{3} +3*(-1)^{2} -1*(-1)-3$

-1+3+1-3=0

$x^{3} +3x^{2} -x-3$

$*(-3)^{3} +3*(-3)^{2} -1*(-3)-3$

-27+27+3-3=0

Verified.

Now,p+q+r= 1-1-3=-3=-b/a

    pqr= 1*(-1)*(-3)=3=-d/a

pq+qr+rp= 1*(-1)+(-1)*(-3)+(-3)*(1)=-1+3-3=1=c/a

Hence,verified

Hope it helps

Thanks

Answer 2

Solution :

Let given cubic polynomial be

p(x) = x³ + 3x² - x - 3

i ) If x = 1 , then

p(1) = 1³ + 3(1)² - 1 - 3 = 1 + 3 - 1 - 3 = 0

ii ) If x = -1 , then

p(-1) = ( -1 )³ + 3( -1 )² - ( -1 ) - 3

= -1 + 3 + 1 - 3

= 0

iii ) If x = -3 , then

p(-3) = (-3)³ + 3(-3)²-(-3)-3

= -27 + 27 + 3 - 3

= 0

Therefore ,

p(1) = p(-1) = p(-3) = 0.

So, 1 , -1 , -3 are zeroes of the given

cubic polynomial p(x).


Compare the coefficients of given

cubic polynomial x³+3x²-x-3 with

ax³+bx²+cx+d , we get

a = 1 , b = 3 , c = -1 , d = -3

Let the zeroes of the cubic polynomial

p(x) are p = 1 , q = -1 , r = -3

Now ,

p + q + r = 1 - 1 - 3 = -3 = -b/a

pq + qr + rp = 1(-1)+(-1)(-3)+(-3)1

= -1 + 3 - 3

= -1 = c/a

pqr = 1(-1)(-3) = 3 = -d/a

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