Let us consider ${p(x) = 2x^3 – 5x^2 – 14x + 8}$.
You can check that p(x) = 0 for x = 4, – 2, ${1/2}$ Since p(x) can have atmost three zeroes,
these are the zeores of ${2x^3 – 5x^2 – 14x + 8}$. Now,
Sum of its zeroes = (4)+ (– 2) + ${1/2}$ = ${5/2}$ = ${-Coefficient of x^2}/{Coefficient of x^3}$
Product of its zeroes = (4)x (– 2) x ${1/2}$ = ${-8/2}$ = ${-Constant term}/{coefficient of x^3}$
However, there is one more relationship here.
Consider the sum of the products of the zeroes taken two at a time. We have:
=(4)x (– 2) + (– 2) x ${1/2}$ + (4) x ${1/2}$
= ${-14/2}$
= ${Coefficient of x}/{Coefficient of x^3}$
In general, it can be proved that if α, β, γ are the zeroes of the cubic polynomial ${ax^3 + bx^2 + cx + d}$ ,
then
Sum of its zeroes = α + β + γ = ${-b/a}$ ,
Sum of the products of the zeroes taken two at a time =αβ + βγ + γα = ${c/a}$ ,
Product of its zeroes = αβγ = ${-d/a}$
The zeroes of a cubic polynomial are ${3}$, ${-1}$, ${-1/3}$. Find the polynomial.
Let the cubic polynomial be ${ax^3 + bx^2 + cx + d}$.
Since,
Sum of its zeroes = α + β + γ = $3 + (-1) + (-1/3)$=${-b/a}$
= $3 -4/3$ ,
= $5/3$ = ${-b/a}$ ,
Sum of the products of the zeroes taken two at a time =αβ + βγ + γα = ${c/a}$
= $3× (-1)+ (-1)×(-1/3) + 3×(-1/3)$ ,
= $-3 +1/3 -1$ ,
= $1/3 -4$ ,
= $-11/3$ = ${c/a}$
Product of its zeroes = αβγ = ${-d/a}$
= $(3)×(-1)×(-1/3)$
= $1$ = ${-d/a}$
${ax^3 + bx^2 + cx + d}$=>${x^3 + b/ax^2 + c/ax + d/a}$
Substituting the values in above equation, we have
= ${x^3 + (-5/3)x^2 + (-11/3)x + (-1)}$
= ${3x^3 -5x^2 -11x -3}$
Ans: Required polynomial is ${3x^3 -5x^2 -11x -3}$
The zeroes of a cubic polynomial are ${3}$, ${0}$, ${-1/3}$. Find the polynomial.
Let the cubic polynomial be ${ax^3 + bx^2 + cx + d}$.
Since,
Sum of its zeroes = α + β + γ = $3 + (0) + (-1/3)$=${-b/a}$
= $(9-1)/3$ ,
= $8/3$ = ${-b/a}$ ,
Sum of the products of the zeroes taken two at a time =αβ + βγ + γα = ${c/a}$
= $3× (0)+ (0)×(-1/3) + 3×(-1/3)$ ,
= $-1$ = ${c/a}$
Product of its zeroes = αβγ = ${-d/a}$
= $(3)×(0)×(-1/3)$
= $0$ = ${-d/a}$
${ax^3 + bx^2 + cx + d}$=>${x^3 + b/ax^2 + c/ax + d/a}$
Substituting the values in above equation, we have
= ${x^3 + (-8/3)x^2 + (-1)x + (0)}$
= ${3x^3 -8x^2 -3x }$
Ans: Required polynomial is ${3x^3 -8x^2 -3x}$
the sum of the products of the zeroes taken two at a time is -1. Find the polynomial.
Let the cubical polynomial be ${ax^3 + bx^2 + cx + d}$
Given:
Sum of its zeroes = α + β + γ = $0$=${-b/a}$
Sum of the products of the zeroes taken two at a time =αβ + βγ + γα = ${c/a}$ ,
= $-1$ = ${c/a}$
Product of its zeroes = αβγ = ${-d/a}$
= $0$ = ${-d/a}$
Therefore the polynomial is,${ax^3 + bx^2 + cx + d}$ = ${x^3 + (0)x^2 + (-1)x + 0}$
=${x^3 -x}$
Ans: Required polynomial is ${x^3 -x}$
k is zeroes of any polynomial ${p(x)}$ if ${p(k)=0}$.
${p(3) = 3 × 3^3 – (5 × 3^2) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,}$
${p(–1) = 3 × (–1)^3 – 5 × (–1)^2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,}$
${ p(–1/3)$ = $3 × (–1/3)^3 – 5 × (–1/3)^2 – 11 × (–1/3) – 3$ = $(-1/9)-(5/9) +(11/3)$ - $3$
= ${-(2/3) + (2/3) = 0 }$
Therefore ${3}$, ${-1}$,${-1/3}$ are the zeroes of the cubic polynomial ${p(x) = 3x^3 – 5x^2 – 11x – 3}$