If zeroes are 2, -1, 3 then polynomial is (x-2)(x+1)(x-3). Expand and verify sum/product of zeroes.

Form a Cubic Polynomial with Zeroes 2, -1, 3 — Constructing Polynomials

Question

Given that the zeroes of a cubic polynomial are $2$ , $-1$ , and $3$ , form the polynomial. Also verify using the relations between zeroes and coefficients.

Solution — Step by Step

If $\alpha$ is a zero of a polynomial $p(x)$ , then $(x - \alpha)$ is a factor. So our three zeroes $2$ , $-1$ , $3$ give us three factors: $(x-2)$ , $(x+1)$ , $(x-3)$ .

The polynomial is their product:

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

We take $(x-2)(x+1)$ first — pairs with opposite signs are easier to handle:

(x-2)(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2

Now multiply $(x^2 - x - 2)$ with $(x - 3)$ :

(x^2 - x - 2)(x - 3) = x^3 - 3x^2 - x^2 + 3x - 2x + 6

= x^3 - 4x^2 + x + 6

So $p(x) = x^3 - 4x^2 + x + 6$ .

For a cubic $ax^3 + bx^2 + cx + d$ with zeroes $\alpha, \beta, \gamma$ :

\alpha + \beta + \gamma = -\frac{b}{a}, \quad \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}, \quad \alpha\beta\gamma = -\frac{d}{a}

Plugging in our values ( $a=1, b=-4, c=1, d=6$ ):

Relation	LHS (zeroes)	RHS (coefficients)
Sum	$2 + (-1) + 3 = 4$	$-(-4)/1 = 4$ ✓
Sum of products (pairs)	$2(-1) + (-1)(3) + (3)(2) = -2 - 3 + 6 = 1$	$1/1 = 1$ ✓
Product	$2 \times (-1) \times 3 = -6$	$-6/1 = -6$ ✓

All three match.

Why This Works

The factor theorem is the backbone here: a polynomial $p(x)$ has zero $\alpha$ if and only if $(x - \alpha)$ divides $p(x)$ exactly. We’re running this in reverse — given the zeroes, we construct the factors.

The coefficient relations (Vieta’s formulas) aren’t just a check — they’re a shortcut. If a question gives you sum and product of zeroes directly, you can write the polynomial without expanding anything.

For boards, this verification step is mandatory to get full marks. Even if your expansion is correct, skipping verification costs you a step’s worth of marks in CBSE marking schemes.

Alternative Method

If a question gives you the sum of zeroes $(\alpha+\beta+\gamma)$ , sum of products of pairs $(\alpha\beta+\beta\gamma+\gamma\alpha)$ , and product $(\alpha\beta\gamma)$ directly, use the standard form:

p(x) = x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma

Here: sum $= 4$ , sum of pairs $= 1$ , product $= -6$ . So:

p(x) = x^3 - 4x^2 + x - (-6) = x^3 - 4x^2 + x + 6

Same answer, zero expansion required. This is the method to use in exams when you’re short on time.

Memorise the sign pattern: $-(\text{sum}),\ +(\text{sum of pairs}),\ -(\text{product})$ . The alternating signs trip students up constantly.

Common Mistake

The most common error is writing $(x+2)(x-1)(x+3)$ — students flip the signs on all three factors. Remember: zero is $2$ means the factor is $(x \mathbf{-} 2)$ , not $(x+2)$ . If you substitute $x = 2$ into $(x+2)$ , you get $4$ , not $0$ . The factor must give zero when you plug the zero in.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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