Polynomials | Exercise 2.2

Question 1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) $x^2 - 2x - 8$

(ii) $4s^2 - 4s + 1$

(iii) $6x^2 - 3 - 7x$

(iv) $4u^2 + 8u$

(v) $t^2 - 15$

(vi) $3x^2 - x - 4$

Solution

Understand the Question

For any quadratic polynomial $p(x) = ax^2 + bx + c$ :

The zeroes $\alpha$ and $\beta$ are the values of $x$ for which $p(x) = 0$ .
The relationships between the zeroes and coefficients are:
- Sum of zeroes: $\alpha + \beta = -\dfrac{b}{a} = -\dfrac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$
- Product of zeroes: $\alpha \cdot \beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Coefficient of } x^2}$
To solve each part, first factorise the polynomial to find its zeroes, then evaluate the sum and product directly and compare them with $-\dfrac{b}{a}$ and $\dfrac{c}{a}$ .

(i) $x^2 - 2x - 8$

Step 1 · Find the Zeroes

To find the zeroes, set $x^2 - 2x - 8 = 0$ : Diagram 1

\begin{aligned} x^2 - 4x + 2x - 8 &= 0 \\ x(x - 4) + 2(x - 4) &= 0 \\ (x - 4)(x + 2) &= 0 \end{aligned}

$x - 4 = 0 \quad \text{or} \quad x + 2 = 0$ $x = 4 \quad \text{or} \quad x = -2$

Therefore, the zeroes are $\alpha = 4$ and $\beta = -2$ .

Step 2 · Verify Relationship with Coefficients

Comparing $x^2 - 2x - 8$ with $ax^2 + bx + c$ , we get $a = 1$ , $b = -2$ , and $c = -8$ .

Sum of zeroes:

\begin{aligned} \alpha + \beta &= 4 + (-2) \\[0.6em] &= 2 \end{aligned}

$-\dfrac{b}{a} = -\dfrac{-2}{1} = 2$

$\alpha + \beta = -\dfrac{b}{a}$

Product of zeroes:

\begin{aligned} \alpha \cdot \beta &= 4 \cdot (-2) \\[0.6em] &= -8 \end{aligned}

$\dfrac{c}{a} = \dfrac{-8}{1} = -8$

$\alpha \cdot \beta = \dfrac{c}{a}$

Answer

(i) Zeroes are $4$ and $-2$ . The relationships $\alpha + \beta = -\dfrac{b}{a} = 2$ and $\alpha \cdot \beta = \dfrac{c}{a} = -8$ are verified.

(ii) $4s^2 - 4s + 1$

Step 1 · Find the Zeroes

To find the zeroes, set $4s^2 - 4s + 1 = 0$ :

\begin{aligned} (2s - 1)^2 &= 0 \\ 2s - 1 &= 0 \\ 2s &= 1 \\ s &= \dfrac{1}{2} \end{aligned}

Therefore, the zeroes are $\alpha = \dfrac{1}{2}$ and $\beta = \dfrac{1}{2}$ .

Step 2 · Verify Relationship with Coefficients

Comparing $4s^2 - 4s + 1$ with $as^2 + bs + c$ , we get $a = 4$ , $b = -4$ , and $c = 1$ .

Sum of zeroes:

\begin{aligned} \alpha + \beta &= \dfrac{1}{2} + \dfrac{1}{2} \\[0.6em] &= 1 \end{aligned}

$-\dfrac{b}{a} = -\dfrac{-4}{4} = 1$

$\alpha + \beta = -\dfrac{b}{a}$

Product of zeroes:

\begin{aligned} \alpha \cdot \beta &= \left(\dfrac{1}{2}\right) \cdot \left(\dfrac{1}{2}\right) \\[0.6em] &= \dfrac{1}{4} \end{aligned}

$\dfrac{c}{a} = \dfrac{1}{4}$

$\alpha \cdot \beta = \dfrac{c}{a}$

Answer

(ii) Zeroes are $\dfrac{1}{2}$ and $\dfrac{1}{2}$ . The relationships $\alpha + \beta = -\dfrac{b}{a} = 1$ and $\alpha \cdot \beta = \dfrac{c}{a} = \dfrac{1}{4}$ are verified.

(iii) $6x^2 - 3 - 7x$

Step 1 · Find the Zeroes

Write in standard form: $6x^2 - 7x - 3 = 0$ .

\begin{aligned} 6x^2 - 9x + 2x - 3 &= 0 \\ 3x(2x - 3) + 1(2x - 3) &= 0 \\ (3x + 1)(2x - 3) &= 0 \end{aligned}

$3x + 1 = 0 \quad \text{or} \quad 2x - 3 = 0$ $3x = -1 \quad \text{or} \quad 2x = 3$ $x = -\dfrac{1}{3} \quad \text{or} \quad x = \dfrac{3}{2}$

Therefore, the zeroes are $\alpha = -\dfrac{1}{3}$ and $\beta = \dfrac{3}{2}$ .

Step 2 · Verify Relationship with Coefficients

Comparing $6x^2 - 7x - 3$ with $ax^2 + bx + c$ , we get $a = 6$ , $b = -7$ , and $c = -3$ .

Sum of zeroes:

\begin{aligned} \alpha + \beta &= -\dfrac{1}{3} + \dfrac{3}{2} \\[0.6em] &= -\dfrac{2}{6} + \dfrac{9}{6} \\[0.6em] &= \dfrac{7}{6} \end{aligned}

$-\dfrac{b}{a} = -\dfrac{-7}{6} = \dfrac{7}{6}$

$\alpha + \beta = -\dfrac{b}{a}$

Product of zeroes:

\begin{aligned} \alpha \cdot \beta &= \left(-\dfrac{1}{3}\right) \cdot \left(\dfrac{3}{2}\right) \\[0.6em] &= -\dfrac{3}{6} \\[0.6em] &= -\dfrac{1}{2} \end{aligned}

$\dfrac{c}{a} = \dfrac{-3}{6} = -\dfrac{1}{2}$

$\alpha \cdot \beta = \dfrac{c}{a}$

Answer

(iii) Zeroes are $-\dfrac{1}{3}$ and $\dfrac{3}{2}$ . The relationships $\alpha + \beta = -\dfrac{b}{a} = \dfrac{7}{6}$ and $\alpha \cdot \beta = \dfrac{c}{a} = -\dfrac{1}{2}$ are verified.

(iv) $4u^2 + 8u$

Step 1 · Find the Zeroes

To find the zeroes, set $4u^2 + 8u = 0$ :

\begin{aligned} 4u(u + 2) &= 0 \end{aligned}

$4u = 0 \quad \text{or} \quad u + 2 = 0$ $u = 0 \quad \text{or} \quad u = -2$

Therefore, the zeroes are $\alpha = 0$ and $\beta = -2$ .

Step 2 · Verify Relationship with Coefficients

Comparing $4u^2 + 8u + 0$ with $au^2 + bu + c$ , we get $a = 4$ , $b = 8$ , and $c = 0$ .

Sum of zeroes:

\begin{aligned} \alpha + \beta &= 0 + (-2) \\[0.6em] &= -2 \end{aligned}

$-\dfrac{b}{a} = -\dfrac{8}{4} = -2$

$\alpha + \beta = -\dfrac{b}{a}$

Product of zeroes:

\begin{aligned} \alpha \cdot \beta &= 0 \cdot (-2) \\[0.6em] &= 0 \end{aligned}

$\dfrac{c}{a} = \dfrac{0}{4} = 0$

$\alpha \cdot \beta = \dfrac{c}{a}$

Answer

(iv) Zeroes are $0$ and $-2$ . The relationships $\alpha + \beta = -\dfrac{b}{a} = -2$ and $\alpha \cdot \beta = \dfrac{c}{a} = 0$ are verified.

(v) $t^2 - 15$

Step 1 · Find the Zeroes

To find the zeroes, set $t^2 - 15 = 0$ using difference of squares $a^2 - b^2 = (a-b)(a+b)$ :

\begin{aligned} t^2 - (\sqrt{15})^2 &= 0 \\ (t - \sqrt{15})(t + \sqrt{15}) &= 0 \end{aligned}

$t - \sqrt{15} = 0 \quad \text{or} \quad t + \sqrt{15} = 0$ $t = \sqrt{15} \quad \text{or} \quad t = -\sqrt{15}$

Therefore, the zeroes are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$ .

Step 2 · Verify Relationship with Coefficients

Comparing $t^2 + 0t - 15$ with $at^2 + bt + c$ , we get $a = 1$ , $b = 0$ , and $c = -15$ .

Sum of zeroes:

\begin{aligned} \alpha + \beta &= \sqrt{15} + (-\sqrt{15}) \\[0.6em] &= 0 \end{aligned}

$-\dfrac{b}{a} = -\dfrac{0}{1} = 0$

$\alpha + \beta = -\dfrac{b}{a}$

Product of zeroes:

\begin{aligned} \alpha \cdot \beta &= (\sqrt{15}) \cdot (-\sqrt{15}) \\[0.6em] &= -15 \end{aligned}

$\dfrac{c}{a} = \dfrac{-15}{1} = -15$

$\alpha \cdot \beta = \dfrac{c}{a}$

Answer

(v) Zeroes are $\sqrt{15}$ and $-\sqrt{15}$ . The relationships $\alpha + \beta = -\dfrac{b}{a} = 0$ and $\alpha \cdot \beta = \dfrac{c}{a} = -15$ are verified.

(vi) $3x^2 - x - 4$

Step 1 · Find the Zeroes

To find the zeroes, set $3x^2 - x - 4 = 0$ :

\begin{aligned} 3x^2 - 4x + 3x - 4 &= 0 \\ x(3x - 4) + 1(3x - 4) &= 0 \\ (x + 1)(3x - 4) &= 0 \end{aligned}

$x + 1 = 0 \quad \text{or} \quad 3x - 4 = 0$ $x = -1 \quad \text{or} \quad x = \dfrac{4}{3}$

Therefore, the zeroes are $\alpha = -1$ and $\beta = \dfrac{4}{3}$ .

Step 2 · Verify Relationship with Coefficients

Comparing $3x^2 - x - 4$ with $ax^2 + bx + c$ , we get $a = 3$ , $b = -1$ , and $c = -4$ .

Sum of zeroes:

\begin{aligned} \alpha + \beta &= -1 + \dfrac{4}{3} \\[0.6em] &= -\dfrac{3}{3} + \dfrac{4}{3} \\[0.6em] &= \dfrac{1}{3} \end{aligned}

$-\dfrac{b}{a} = -\dfrac{-1}{3} = \dfrac{1}{3}$

$\alpha + \beta = -\dfrac{b}{a}$

Product of zeroes:

\begin{aligned} \alpha \cdot \beta &= (-1) \cdot \left(\dfrac{4}{3}\right) \\[0.6em] &= -\dfrac{4}{3} \end{aligned}

$\dfrac{c}{a} = \dfrac{-4}{3} = -\dfrac{4}{3}$

$\alpha \cdot \beta = \dfrac{c}{a}$

Answer

(vi) Zeroes are $-1$ and $\dfrac{4}{3}$ . The relationships $\alpha + \beta = -\dfrac{b}{a} = \dfrac{1}{3}$ and $\alpha \cdot \beta = \dfrac{c}{a} = -\dfrac{4}{3}$ are verified.

Common Mistakes

Not Rearranging to Standard Form: In part (iii), $6x^2 - 3 - 7x$ must be rearranged to $6x^2 - 7x - 3$ before identifying coefficients; otherwise $b$ and $c$ are swapped.
Sign Errors in $-\dfrac{b}{a}$ : Forgetting the negative sign in $-\dfrac{b}{a}$ , especially when $b$ is already negative (e.g. $-\dfrac{-2}{1} = +2$ ).
Missing Terms: In part (iv), the constant term is $c = 0$ , and in part (v), the linear term coefficient is $b = 0$ .

Question 1

More questions in Exercise 2.2