Polynomials Class 10: Zeroes, Division Algorithm & Relationship Between Zeroes

The zeroes of a polynomial $p(x)$ are the $x$-coordinates of the points where the graph of $y = p(x)$ intersects the $x$-axis.

Linear polynomial $ax + b$: The graph is a straight line. It intersects the $x$-axis at exactly one point $x = -\frac{b}{a}$, which is the only zero.

Quadratic polynomial $ax^2 + bx + c$: The graph is a parabola. It can intersect the $x$-axis at:
- Two points — two distinct real zeroes (discriminant $> 0$)
- One point — two equal real zeroes (discriminant $= 0$, parabola just touches the axis)
- No points — no real zeroes (discriminant $< 0$, parabola doesn't reach the axis)

The parabola opens upward if $a > 0$ and downward if $a < 0$.

For $p(x) = ax^2 + bx + c$ with zeroes $\alpha$ and $\beta$:

How to remember: Think of it as "$-b/a$" for sum and "$c/a$" for product. The sum has a negative sign; the product doesn't.

Derivation: Since $\alpha$ and $\beta$ are zeroes:

Comparing with $ax^2 + bx + c$:

For $p(x) = ax^3 + bx^2 + cx + d$ with zeroes $\alpha$, $\beta$, and $\gamma$:

Pattern to remember: The signs alternate: $-b/a$, $+c/a$, $-d/a$. The numerator cycles through the coefficients (after the leading one), and the sign alternates starting with negative.

Problem: Find the sum and product of zeroes of $p(x) = 3x^2 - 5x + 2$.

Solution:
Here $a = 3$, $b = -5$, $c = 2$.

Verification: The zeroes of $3x^2 - 5x + 2 = 0$ are $x = 1$ and $x = \frac{2}{3}$.
- Sum: $1 + \frac{2}{3} = \frac{5}{3}$ $\checkmark$
- Product: $1 \times \frac{2}{3} = \frac{2}{3}$ $\checkmark$

Problem: Verify the relationship between zeroes and coefficients for $p(x) = 2x^3 + 3x^2 - 11x - 6$, given that its zeroes are $2, -3, -\frac{1}{2}$.

Solution:
Here $a = 2$, $b = 3$, $c = -11$, $d = -6$.

Sum of zeroes:

Sum of products taken two at a time:

Product of zeroes:

Problem: Find a quadratic polynomial whose zeroes are $3 + \sqrt{2}$ and $3 - \sqrt{2}$.

Solution:
Sum of zeroes $= (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$

Product of zeroes $= (3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7$

The polynomial is:

Answer: $p(x) = x^2 - 6x + 7$ (or any non-zero scalar multiple of it).

Problem: If one zero of $2x^2 - 5x + k$ is the reciprocal of the other, find the value of $k$.

Solution:
Let the zeroes be $\alpha$ and $\frac{1}{\alpha}$.

Product of zeroes:

Answer: $k = 2$.

Problem: If $\alpha$ and $\beta$ are zeroes of $x^2 - 4x + 3$, find a quadratic polynomial whose zeroes are $2\alpha + \beta$ and $\alpha + 2\beta$.

Solution:
From $x^2 - 4x + 3$: $\alpha + \beta = 4$ and $\alpha\beta = 3$.

New zeroes are $2\alpha + \beta$ and $\alpha + 2\beta$.

Sum of new zeroes:

Product of new zeroes:

The required polynomial:

Answer: $p(x) = x^2 - 12x + 35$.

Problem: Divide $p(x) = 3x^3 + x^2 + 2x + 5$ by $g(x) = x^2 + 2x + 1$.

Solution:

Step 1: $\frac{3x^3}{x^2} = 3x$. First term of quotient: $3x$.

$3x \times (x^2 + 2x + 1) = 3x^3 + 6x^2 + 3x$

Subtract: $(3x^3 + x^2 + 2x + 5) - (3x^3 + 6x^2 + 3x) = -5x^2 - x + 5$

Step 2: $\frac{-5x^2}{x^2} = -5$. Next term of quotient: $-5$.

$-5 \times (x^2 + 2x + 1) = -5x^2 - 10x - 5$

Subtract: $(-5x^2 - x + 5) - (-5x^2 - 10x - 5) = 9x + 10$

Since degree of $9x + 10$ (which is 1) $<$ degree of $x^2 + 2x + 1$ (which is 2), we stop.

Verification: $g(x) \times q(x) + r(x) = (x^2 + 2x + 1)(3x - 5) + 9x + 10 = 3x^3 + x^2 + 2x + 5 = p(x)$ $\checkmark$

Problem: Find all zeroes of $2x^4 - 3x^3 - 3x^2 + 6x - 2$, given that two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$.

Solution:
Since $\sqrt{2}$ and $-\sqrt{2}$ are zeroes, $(x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2$ is a factor.

Divide $2x^4 - 3x^3 - 3x^2 + 6x - 2$ by $x^2 - 2$:

Performing polynomial long division:

Now find the zeroes of $2x^2 - 3x + 1$:

Answer: All four zeroes are $\sqrt{2}, -\sqrt{2}, \frac{1}{2}, 1$.

Problem: If the sum of zeroes of $p(x) = kx^2 + 2x + 3k$ is equal to their product, find the value of $k$.

Solution:
Sum of zeroes $= -\frac{2}{k}$

Product of zeroes $= \frac{3k}{k} = 3$

Given: Sum $=$ Product:

Answer: $k = -\frac{2}{3}$.

Problem: Find a quadratic polynomial whose zeroes are $\frac{3}{5}$ and $-\frac{1}{2}$.

Solution:
Sum $= \frac{3}{5} + \left(-\frac{1}{2}\right) = \frac{6 - 5}{10} = \frac{1}{10}$

Product $= \frac{3}{5} \times \left(-\frac{1}{2}\right) = -\frac{3}{10}$

Polynomial $= x^2 - \frac{1}{10}x - \frac{3}{10}$

Multiplying by 10 to clear fractions:

Answer: $p(x) = 10x^2 - x - 3$.

Problem: Check whether $g(x) = x^2 - 3$ is a factor of $p(x) = 2x^4 + 3x^3 - 2x^2 - 9x - 12$.

Solution:
Divide $p(x)$ by $g(x)$:

Remainder $= 0$.

Since the remainder is $0$, $g(x)$ is indeed a factor of $p(x)$. $\checkmark$

Answer: Yes, $x^2 - 3$ is a factor.