Solved Examples

How to Use This Section

This is your practice powerhouse for Polynomials. Below are 30+ fully worked problems spanning the whole chapter — identifying polynomials, finding zeroes, the zero–coefficient relations for quadratics and cubics, forming polynomials, and polynomial division — arranged roughly from easy to hard.

How to study: Try each problem yourself with the solution covered, then check the steps. In the board exam, marks are awarded step by step, so practise writing full solutions.

Keep these formulae handy:

• Quadratic $ax^2+bx+c$: sum of zeroes $=-\dfrac{b}{a}$, product $=\dfrac{c}{a}$.
• Cubic $ax^3+bx^2+cx+d$: $\sum\alpha = -\dfrac{b}{a}$, $\sum\alpha\beta = \dfrac{c}{a}$, $\alpha\beta\gamma = -\dfrac{d}{a}$.
• Quadratic from zeroes: $x^2 - (\text{sum})x + (\text{product})$.
• $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
• Division: $p(x) = g(x)q(x) + r(x)$.

Solved Examples

Example 1: Degree and type

State the degree and type of $p(x) = 5 - 2x + 3x^2$.

Solution:

1. Highest power of $x$ is $x^2$, so degree $= 2$.
2. Degree 2 ⇒ quadratic.

Final Answer: Degree 2, quadratic.

Takeaway: Identify the highest power first.

Example 2: Find the zeroes by factorisation

Find the zeroes of $p(x) = x^2 - 2x - 8$.

Solution:

1. Factorise: $x^2 - 4x + 2x - 8 = x(x-4) + 2(x-4) = (x+2)(x-4)$.
2. Zeroes: $x = -2$ and $x = 4$.

Final Answer: $-2$ and $4$.

Takeaway: Split the middle term to factorise.

Example 3: Sum and product check

Verify the zero–coefficient relations for $p(x) = x^2 - 2x - 8$.

Solution:

1. Zeroes (from Example 2): $-2, 4$. Sum $= 2$, product $= -8$.
2. $-\dfrac{b}{a} = -\dfrac{-2}{1} = 2$. ✓
3. $\dfrac{c}{a} = \dfrac{-8}{1} = -8$. ✓

Final Answer: Both relations verified.

Takeaway: Compare computed sum/product with $-b/a$ and $c/a$.

Example 4: Zeroes of $4x^2 - 9$

Find the zeroes of $p(x) = 4x^2 - 9$.

Solution:

1. This is a difference of squares: $4x^2 - 9 = (2x-3)(2x+3)$.
2. Zeroes: $x = \dfrac{3}{2}$ and $x = -\dfrac{3}{2}$.

Final Answer: $\dfrac{3}{2}$ and $-\dfrac{3}{2}$.

Takeaway: Recognise $a^2 - b^2 = (a-b)(a+b)$.

Example 5: Quadratic from sum and product

Find a quadratic polynomial whose zeroes have sum 5 and product 6.

Solution:

1. $p(x) = x^2 - (\text{sum})x + (\text{product})$.
2. $= x^2 - 5x + 6$.

Final Answer: $x^2 - 5x + 6$.

Takeaway: Use the standard template directly.

Example 6: Find $\alpha^2 + \beta^2$

If $\alpha, \beta$ are zeroes of $x^2 - 4x + 3$, find $\alpha^2 + \beta^2$.

Solution:

1. Sum $= 4$, product $= 3$.
2. $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 16 - 6 = 10$.

Final Answer: 10.

Takeaway: Convert to sum and product.

Example 7: Find $k$ from a given zero

If $x = 2$ is a zero of $p(x) = x^2 + kx - 6$, find $k$.

Solution:

1. $p(2) = 0$: $4 + 2k - 6 = 0$.
2. $2k - 2 = 0 \Rightarrow k = 1$.

Final Answer: $k = 1$.

Takeaway: Substitute the zero and solve for the unknown.

Example 8: Quadratic from irrational zeroes

Form a quadratic whose zeroes are $3 + \sqrt{5}$ and $3 - \sqrt{5}$.

Solution:

1. Sum $= 6$, product $= 9 - 5 = 4$.
2. $p(x) = x^2 - 6x + 4$.

Final Answer: $x^2 - 6x + 4$.

Takeaway: Conjugate surds give rational sum and product.

Example 9: Find $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$

If $\alpha, \beta$ are zeroes of $3x^2 - 5x - 2$, find $\dfrac{1}{\alpha} + \dfrac{1}{\beta}$.

Solution:

1. Sum $= \dfrac{5}{3}$, product $= -\dfrac{2}{3}$.
2. $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{5/3}{-2/3} = -\dfrac{5}{2}$.

Final Answer: $-\dfrac{5}{2}$.

Takeaway: $\dfrac{\text{sum}}{\text{product}}$ gives the sum of reciprocals.

Example 10: Zeroes of $6x^2 - x - 2$

Find the zeroes of $p(x) = 6x^2 - x - 2$.

Solution:

1. Split: $6x^2 - 4x + 3x - 2 = 2x(3x-2) + 1(3x-2) = (2x+1)(3x-2)$.
2. Zeroes: $x = -\dfrac{1}{2}$ and $x = \dfrac{2}{3}$.

Final Answer: $-\dfrac{1}{2}$ and $\dfrac{2}{3}$.

Takeaway: Product of $a\cdot c = -12$, split $-1$ as $-4 + 3$.

Example 11: Quadratic from given zeroes

Form a quadratic with zeroes $-3$ and 5.

Solution:

1. Sum $= 2$, product $= -15$.
2. $p(x) = x^2 - 2x - 15$.

Final Answer: $x^2 - 2x - 15$.

Takeaway: Template $x^2 - (\text{sum})x + (\text{product})$.

Example 12: Cubic relation — sum of zeroes

Find the sum of the zeroes of $p(x) = 2x^3 - 6x^2 + 3x - 1$.

Solution:

1. Sum $= -\dfrac{b}{a} = -\dfrac{-6}{2} = 3$.

Final Answer: 3.

Takeaway: Sum of cubic's zeroes $= -b/a$.

Example 13: Cubic — product of zeroes

Find the product of the zeroes of $p(x) = x^3 - 4x^2 + 5x - 2$.

Solution:

1. Product $= -\dfrac{d}{a} = -\dfrac{-2}{1} = 2$.

Final Answer: 2.

Takeaway: Product of cubic's zeroes $= -d/a$.

Example 14: Form a cubic from zeroes

Find a cubic polynomial with zeroes 0, 2, and $-3$.

Solution:

1. $S_1 = 0 + 2 + (-3) = -1$.
2. $S_2 = (0)(2) + (2)(-3) + (-3)(0) = -6$.
3. $S_3 = 0 \cdot 2 \cdot (-3) = 0$.
4. $p(x) = x^3 - (-1)x^2 + (-6)x - 0 = x^3 + x^2 - 6x$.

Final Answer: $x^3 + x^2 - 6x$.

Takeaway: A zero of 0 makes the constant term 0.

Example 15: Find the third zero

If two zeroes of $x^3 - 6x^2 + 11x - 6$ are 1 and 2, find the third.

Solution:

1. Sum of zeroes $= -\dfrac{-6}{1} = 6$.
2. Third zero $= 6 - (1 + 2) = 3$.

Final Answer: 3.

Takeaway: Use the sum relation to recover the missing zero.

Example 16: Polynomial division (linear divisor)

Divide $p(x) = x^2 + 4x + 4$ by $x + 2$.

Solution:

1. $x^2 + 4x + 4 = (x+2)^2 = (x+2)(x+2)$.
2. Quotient $x + 2$, remainder 0.

Final Answer: Quotient $x + 2$, remainder 0.

Takeaway: A perfect square trinomial divides exactly.

Example 17: Division with remainder

Divide $p(x) = 2x^3 + 3x^2 - x + 1$ by $x + 1$ using the Remainder Theorem to find the remainder.

Solution:

1. Remainder $= p(-1) = 2(-1)^3 + 3(-1)^2 - (-1) + 1$.
2. $= -2 + 3 + 1 + 1 = 3$.

Final Answer: Remainder $= 3$.

Takeaway: Dividing by $(x - a)$ leaves remainder $p(a)$; here $a = -1$.

Example 18: Find all zeroes given two (surds)

Find all zeroes of $p(x) = x^4 - x^3 - 7x^2 + x + 6$, given that $1$ and $-1$ are zeroes.

Solution:

1. $(x-1)(x+1) = x^2 - 1$ is a factor.
2. Dividing gives quotient $x^2 - x - 6 = (x-3)(x+2)$.
3. So remaining zeroes are $3$ and $-2$.

Final Answer: $1, -1, 3, -2$.

Takeaway: Divide by the known factor, then factor the quotient.

Example 19: Find $k$ so a number is a zero

Find $k$ if $-2$ is a zero of $p(x) = 2x^3 + x^2 + kx + 2$.

Solution:

1. $p(-2) = 0$: $2(-8) + 4 - 2k + 2 = 0$.
2. $-16 + 4 - 2k + 2 = 0 \Rightarrow -10 - 2k = 0 \Rightarrow k = -5$.

Final Answer: $k = -5$.

Takeaway: Substitute the zero and solve for the parameter.

Example 20: $\alpha^3 + \beta^3$

If $\alpha, \beta$ are zeroes of $x^2 - 3x + 1$, find $\alpha^3 + \beta^3$.

Solution:

1. Sum $= 3$, product $= 1$.
2. $\alpha^3 + \beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = 27 - 3(1)(3) = 27 - 9 = 18$.

Final Answer: 18.

Takeaway: Use $\alpha^3+\beta^3 = s^3 - 3ps$.

Example 21: Finding $k$ from a sum and product relation

Find $k$ if the sum of the zeroes of $x^2 - (k+6)x + 2(2k-1)$ is half their product.

Solution:

1. Sum $= k + 6$; product $= 2(2k-1) = 4k - 2$.
2. Condition: sum $= \dfrac{1}{2}\times$ product ⇒ $k + 6 = \dfrac{1}{2}(4k - 2) = 2k - 1$.
3. $6 + 1 = 2k - k \Rightarrow k = 7$.

Final Answer: $k = 7$.

Takeaway: Translate the worded relation into an equation in sum and product.

Example 22: Verify division algorithm

Divide $p(x) = x^3 + x^2 + x + 1$ by $x + 1$ and verify.

Solution:

1. Group: $x^2(x+1) + 1(x+1) = (x+1)(x^2 + 1)$.
2. So quotient $= x^2 + 1$, remainder 0.
3. Check: $(x+1)(x^2+1) = x^3 + x^2 + x + 1$. ✓

Final Answer: Quotient $x^2 + 1$, remainder 0.

Takeaway: Factor by grouping when possible, then verify.

Example 23: Number of zeroes from a graph

A quadratic polynomial's graph touches the x-axis at exactly one point. How many distinct zeroes does it have, and what does this say about the zeroes?

Solution:

1. Touching at one point ⇒ one distinct zero.
2. The two zeroes are equal (a repeated/double root).

Final Answer: One distinct zero; the zeroes are equal.

Takeaway: Touching the axis = equal zeroes.

Example 24: Quadratic with equal zeroes — find $k$

For what value of $k$ does $x^2 - kx + 9$ have equal zeroes (touching parabola)?

Solution:

1. Equal zeroes mean both zeroes are the same, say $\alpha$. Then product $= \alpha^2 = 9$, so $\alpha = \pm 3$.
2. Sum $= 2\alpha = k$. So $k = 6$ (for $\alpha = 3$) or $k = -6$ (for $\alpha = -3$).

Final Answer: $k = 6$ or $k = -6$.

Takeaway: Equal zeroes ⇒ product is a perfect square of the repeated root; sum is twice it.

Example 25: Sum of squares for a cubic application

If $\alpha, \beta$ are zeroes of $x^2 + 5x + 6$, form a polynomial whose zeroes are $\alpha + 1$ and $\beta + 1$.

Solution:

1. Original sum $= -5$, product $= 6$.
2. New sum $= (\alpha+1)+(\beta+1) = (\alpha+\beta) + 2 = -5 + 2 = -3$.
3. New product $= (\alpha+1)(\beta+1) = \alpha\beta + (\alpha+\beta) + 1 = 6 - 5 + 1 = 2$.
4. Polynomial: $x^2 - (-3)x + 2 = x^2 + 3x + 2$.

Final Answer: $x^2 + 3x + 2$.

Takeaway: Shift the zeroes by computing the new sum and product.

Example 26: Find a zero of a cubic by inspection

Show that $x = 1$ is a zero of $p(x) = x^3 - 6x^2 + 11x - 6$ and find the others.

Solution:

1. $p(1) = 1 - 6 + 11 - 6 = 0$, so 1 is a zero; $(x-1)$ is a factor.
2. Divide: quotient $x^2 - 5x + 6 = (x-2)(x-3)$.
3. Other zeroes: 2 and 3.

Final Answer: Zeroes $1, 2, 3$.

Takeaway: Test small integer values to find a first zero, then factor.

Example 27: Quadratic with zeroes negatives of another

If $\alpha, \beta$ are zeroes of $x^2 - 7x + 10$, form a polynomial whose zeroes are $-\alpha$ and $-\beta$.

Solution:

1. Original sum $= 7$, product $= 10$.
2. New sum $= -\alpha - \beta = -(\alpha+\beta) = -7$.
3. New product $= (-\alpha)(-\beta) = \alpha\beta = 10$.
4. Polynomial: $x^2 - (-7)x + 10 = x^2 + 7x + 10$.

Final Answer: $x^2 + 7x + 10$.

Takeaway: Negating zeroes flips the sign of the middle term.

Example 28: Division by a quadratic with remainder

Divide $p(x) = x^3 + 2x^2 + 3$ by $x^2 + x + 1$.

Solution:

1. $\dfrac{x^3}{x^2} = x$; $x(x^2+x+1) = x^3 + x^2 + x$; subtract: $(x^3 + 2x^2 + 0x) - (x^3+x^2+x) = x^2 - x$. Bring down $+3$: $x^2 - x + 3$.
2. $\dfrac{x^2}{x^2} = 1$; $1(x^2+x+1) = x^2 + x + 1$; subtract: $(x^2 - x + 3) - (x^2 + x + 1) = -2x + 2$.
3. Degree 1 < 2, stop. Quotient $x + 1$, remainder $-2x + 2$.

Final Answer: Quotient $x + 1$, remainder $-2x + 2$.

Takeaway: Insert $0x$ for the missing $x$-term before dividing.

Example 29: Find both zeroes' relationship

If the zeroes of $x^2 - px + q$ are in the ratio 2 : 3, and the product is 24, find the zeroes.

Solution:

1. Let the zeroes be $2t$ and $3t$. Product $= 6t^2 = 24 \Rightarrow t^2 = 4 \Rightarrow t = 2$ (taking positive).
2. Zeroes $= 4$ and $6$.

Final Answer: The zeroes are 4 and 6.

Takeaway: Use the ratio to parametrise the zeroes, then apply the product.

Example 30: Forming a cubic from a condition

Find a cubic polynomial whose zeroes are 2, 2, and 3 (note the repeated zero).

Solution:

1. $S_1 = 2 + 2 + 3 = 7$.
2. $S_2 = (2)(2) + (2)(3) + (3)(2) = 4 + 6 + 6 = 16$.
3. $S_3 = 2 \cdot 2 \cdot 3 = 12$.
4. $p(x) = x^3 - 7x^2 + 16x - 12$.

Final Answer: $x^3 - 7x^2 + 16x - 12$.

Takeaway: Repeated zeroes are counted as many times as they occur in $S_1, S_2, S_3$.

Example 31: Find $\alpha - \beta$ (positive value)

If $\alpha, \beta$ are zeroes of $x^2 - 7x + 12$ with $\alpha > \beta$, find $\alpha - \beta$.

Solution:

1. Sum $= 7$, product $= 12$.
2. $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = 49 - 48 = 1$.
3. So $\alpha - \beta = 1$ (positive, since $\alpha > \beta$).

Final Answer: $\alpha - \beta = 1$.

Takeaway: Find $(\alpha-\beta)^2$, then take the positive square root.