3.6 CUBIC POLYNOMIALS

Let us now look at cubic polynomials. Do you think similar relation holds between the zeroes of a cubic polynomial and its coefficients as well?

Let us consider p(x) = 2x 4– 5x 2 – 14x + 8.

We see that p(x) = 0 for x = 4, – 2, 1/ 2 .

Since p(x) can have at most three zeroes, these are the zeroes of 2x3 – 5x 22 – 14x + 8.

Sum of its zeroes = 4 + (–2) + 1/ 2 = 5/ 2 = -(-5) 2/ = (coefficient of x2)/ coefficient of x3

Product of its zeroes = 4 × (–2) × 1/ 2 = – 4 = -8/ 2 = 3 -(constant term)/ coefficientof x3 ,

However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have:

=

ax3 + bx2+ cx + d is a polynomial with zeroes abg , , . Let us see how αβy , , relate to a, b, c, d.

Since αβy are the zeroes, the polynomial can be written as (x - α)( x- β)( x - y )

= x3 - x2 (α+β+ y) + x(αβ + βy+ yα) - αβy

To compare with the polynomial, we multiply by 'a' and get

ax3 - x2a (α+β+ y) + xa(αβ + βy+ yα) - aαβy

Therefore;b = -a (α+β+ y),c = a(αβ + βy+ yα),d = -aαβy


{4x(-3)}+{(-2)x1/2}+{1/2 x4}

= – 8 – 1 + 2 = – 7 = -14/ 2 = constant of x/ coefficient of x3

In general, it can be proved that if α,β,y are the zeroes of the cubic polynomial

ax3 + bx2+ cx + d,

α+β+ y= -b/a ,

αβ + βy+ yα = c/ a

and αβy = -d/a.



page no:69

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