Relationship Between Zeroes and Coefficients in Cubic Polynomial

 

❓ Question

Verify that:

$$3,\ -1,\ -\frac{1}{3}$$

are the zeroes of the cubic polynomial:

$$p(x)=3x^3-5x^2-11x-3$$

and verify the relationship between the zeroes and the coefficients.

---

🖼️ Solution Image

---

✍️ Short Explanation

For a cubic polynomial:

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

if the zeroes are:

$$\alpha,\beta,\gamma$$

then:

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

💯

---

🔹 Step 1 — Write the Polynomial

$$p(x)=3x^3-5x^2-11x-3$$

Comparing with:

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

we get:

$$a=3,\quad b=-5,\quad c=-11,\quad d=-3$$

---

🔹 Step 2 — Verify the Given Zeroes

(i) For $x=3$

$$p(3)=3(3)^3-5(3)^2-11(3)-3$$
$$=81-45-33-3$$
$$=0$$

So, $3$ is a zero.

---

(ii) For $x=-1$

$$p(-1)=3(-1)^3-5(-1)^2-11(-1)-3$$
$$=-3-5+11-3$$
$$=0$$

So, $-1$ is a zero.

---

(iii) For $x=-\frac13$

$$p\left(-\frac13\right)=3\left(-\frac13\right)^3-5\left(-\frac13\right)^2-11\left(-\frac13\right)-3$$
$$=-\frac19-\frac59+\frac{11}{3}-3$$
$$=0$$

So,

$$-\frac13$$

is also a zero.

---

🔹 Step 3 — Verify Relationship Between Zeroes and Coefficients

Let:

$$\alpha=3,\quad \beta=-1,\quad \gamma=-\frac13$$

---

(i) Sum of Zeroes

$$\alpha+\beta+\gamma$$
$$=3+(-1)+\left(-\frac13\right)$$
$$=\frac53$$

Also,

$$-\frac{b}{a}=-\frac{-5}{3}=\frac53$$

Verified ✅

---

(ii) Sum of Product of Zeroes Taken Two at a Time

$$\alpha\beta+\beta\gamma+\gamma\alpha$$
$$=3(-1)+(-1)\left(-\frac13\right)+\left(-\frac13\right)(3)$$
$$=-3+\frac13-1$$
$$=-\frac{11}{3}$$

Also,

$$\frac{c}{a}=\frac{-11}{3}$$

Verified ✅

---

(iii) Product of Zeroes

$$\alpha\beta\gamma$$
$$=3(-1)\left(-\frac13\right)$$
$$=1$$

Also,

$$-\frac{d}{a}=-\frac{-3}{3}=1$$

Verified ✅

---

✅ Final Answer

The given numbers:

$$\boxed{3,\ -1,\ -\frac13}$$

are zeroes of the polynomial:

$$\boxed{3x^3-5x^2-11x-3}$$

and all relationships between zeroes and coefficients are verified.

---

⭐ Key Insight

For cubic polynomial:

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

• Sum of zeroes:

$$-\frac{b}{a}$$

• Sum of products of zeroes taken two at a time:

$$\frac{c}{\mathrm{a}}$$

• Product of zeroes:

$$-\frac{d}{a}$$

🧠 Memory Line:

Cubic polynomial me signs alternate hote hain:
$-b/a,\ c/a,\ -d/a$

---

📚 Related Topics

• Prove that 3 + 2√5 is Irrational
• Proof that √5 is Irrational
• Prove that Given Numbers are Irrational