For ax3+ bx2+ cx + d with roots r1, r2, r… | Answer

Subject

গণিত

Topic

Algebra

ক)

Σr_i = b/a; Σ r_ir_j = −c/a; r1r2r3 = d/a

খ)

Σr_i = −b/a; Σ r_ir_j = c/a; r1r2r3 = −d/a

গ)

Σr_i = −c/a; Σ r_ir_j = b/a; r1r2r3 = −d/a

ঘ)

Σr_i = −b/a; Σ r_ir_j = −c/a; r1r2r3 = d/a

We start from the factorization by the roots. If the cubic

ax^3+bx^2+cx+d

has roots

r_1,r_2,r_3

, then (up to the leading coefficient)

ax^3+bx^2+cx+d = a(x-r_1)(x-r_2)(x-r_3).

Expand the right-hand side:

(x-r_1)(x-r_2)(x-r_3)=x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3.

Multiplying by

a

gives

ax^3 - a(r_1+r_2+r_3)x^2 + a(r_1r_2+r_1r_3+r_2r_3)x - a\,r_1r_2r_3.

Compare coefficients with

ax^3+bx^2+cx+d

: - Coefficient of

x^2

\;b = -a(r_1+r_2+r_3)\;

\;r_1+r_2+r_3 = -\dfrac{b}{a}.

- Coefficient of

x

\;c = a(r_1r_2+r_1r_3+r_2r_3)\;

\;r_1r_2+r_1r_3+r_2r_3 = \dfrac{c}{a}.

- Constant term:

\;d = -a\,r_1r_2r_3\;

\;r_1r_2r_3 = -\dfrac{d}{a}.

Therefore the correct Vieta relations are

\boxed{\,r_1+r_2+r_3 = -\dfrac{b}{a},\qquad r_1r_2+r_1r_3+r_2r_3 = \dfrac{c}{a},\qquad r_1r_2r_3 = -\dfrac{d}{a}\,}

This corresponds to Option 2. The alternative list you gave (with the pairwise sum

-c/a

and product

d/a

) has the signs reversed and is not correct for the polynomial written as

ax^3+bx^2+cx+d

For ax3+ bx2+ cx + d with roots r1, r2, r3, state Vieta’s relations.