Polynomials Practice Questions (Part 2)

This lesson works through a set of Class X polynomial questions: finding degrees and zeros, building quadratic polynomials, evaluating expressions in the zeros, and factorising quadratics with a clean four-case method.

Degree of a polynomial

The degree is the highest power of $x$ that appears. For

$3x^5 - 4x^4 + 8x - 5$

the highest power is $5$, so the degree is $5$.

Checking whether a value is a zero

To check if $x = 2$ is a zero of $2x^2 - 4x$, substitute it in:

$2(2)^2 - 4(2) = 2(4) - 4(2) = 8 - 8 = 0.$

Since the value is $0$, $x = 2$ is a zero of $2x^2 - 4x$.

Forming a quadratic from its zeros

If $\alpha$ and $\beta$ are the zeros, the quadratic is

$x^2 - (\alpha + \beta)x + \alpha\beta.$

Zeros $2$ and $3$. Here $\alpha + \beta = 2 + 3 = 5$ and $\alpha\beta = 2 \times 3 = 6$, so

$x^2 - 5x + 6.$

Sum of zeros $10$, product $16$. Substituting the sum and product directly,

$x^2 - 10x + 16.$

Reciprocal zeros: finding $k$

For $3x^2 - 10x + (2k - 1)$, one zero is the reciprocal of the other, so the zeros are $\alpha$ and $\tfrac{1}{\alpha}$. Their product is

$\alpha \cdot \tfrac{1}{\alpha} = 1.$

The product of the zeros also equals $\tfrac{c}{a} = \tfrac{2k - 1}{3}$. Setting them equal,

$\tfrac{2k - 1}{3} = 1 \implies 2k - 1 = 3 \implies 2k = 4 \implies k = 2.$

Finding zeros by splitting the middle term

To find the zeros of $\sqrt{3}\,x^2 + 10x + 7\sqrt{3}$, set the polynomial to $0$. We need two numbers with product $\sqrt{3} \times 7\sqrt{3} = 21$ and sum $10$; these are $3$ and $7$. Splitting the middle term,

$\sqrt{3}\,x^2 + 3x + 7x + 7\sqrt{3} = 0.$

$\sqrt{3}\,x(x + \sqrt{3}) + 7(x + \sqrt{3}) = 0.$

$(\sqrt{3}\,x + 7)(x + \sqrt{3}) = 0.$

So $x = -\sqrt{3}$ or $x = -\tfrac{7}{\sqrt{3}}$. The zeros are $-\sqrt{3}$ and $-\tfrac{7}{\sqrt{3}}$.

Using sum and difference of zeros

For $x^2 - 7x + k$ with zeros $\alpha$ and $\beta$, we are told $\alpha - \beta = 3$. From the polynomial,

$\alpha + \beta = -\tfrac{b}{a} = 7.$

Solving the two equations together:

$(\alpha + \beta) + (\alpha - \beta) = 7 + 3 \implies 2\alpha = 10 \implies \alpha = 5,$

and then $\beta = 7 - 5 = 2$. Since $k = \alpha\beta$,

$k = 5 \times 2 = 10.$

Expressions in the zeros

An expression with $P$ and $Q$. If $P$ and $Q$ are the zeros of $f(t) = t^2 - 4t + 3$, then

$P + Q = -\tfrac{b}{a} = 4, \qquad PQ = \tfrac{c}{a} = 3.$

Evaluate $\tfrac{1}{P} + \tfrac{1}{Q} - 2PQ$ by combining the fractions:

$\frac{1}{P} + \frac{1}{Q} - 2PQ = \frac{P + Q}{PQ} - 2PQ = \frac{4}{3} - 2(3) = \frac{4}{3} - 6 = \frac{4 - 18}{3} = -\frac{14}{3}.$

Sum of squares of the zeros. For $9x^2 - 22x + 8$ with zeros $\alpha$ and $\beta$,

$\alpha + \beta = \tfrac{22}{9}, \qquad \alpha\beta = \tfrac{8}{9}.$

Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,

$\alpha^2 + \beta^2 = \left(\tfrac{22}{9}\right)^2 - 2 \cdot \tfrac{8}{9} = \frac{484}{81} - \frac{16}{9} = \frac{484 - 144}{81} = \frac{340}{81}.$

Rectangle area: length and breadth

The area of a rectangle is $x^2 - 9x + 18$ square units. Factorise to find the sides: we need two numbers with product $18$ and sum $-9$, namely $-6$ and $-3$. So

$x^2 - 9x + 18 = (x - 6)(x - 3),$

giving length $x - 6$ and breadth $x - 3$.

Four cases for factorising a quadratic

To write $x^2 + (a + b)x + ab$ as $(x + a)(x + b)$, look at the signs of the product (the constant times the leading coefficient) and the sum (the coefficient of $x$).

• Product positive, sum positive: both numbers positive; add the numbers.
• Product positive, sum negative: both numbers negative; add the numbers.
• Product negative, sum positive: numbers have opposite signs, the larger positive; take their difference.
• Product negative, sum negative: numbers have opposite signs, the larger negative; take their difference.

Worked factorising examples

When the coefficient of $x^2$ is $1$, the two numbers can be written straight into the factors.

$x^2 + 8x + 15$. Product $15$, sum $8$: both positive, $5$ and $3$. So

$x^2 + 8x + 15 = (x + 5)(x + 3),$

with zeros $x = -5$ and $x = -3$.

$x^2 - 8x + 15$. Product $15$, sum $-8$: both negative, $-5$ and $-3$. So

$x^2 - 8x + 15 = (x - 5)(x - 3),$

with zeros $x = 5$ and $x = 3$.

$x^2 - 5x - 50$. Product $-50$, sum $-5$: opposite signs, larger negative; product $50$, difference $5$ gives $10$ and $5$, so $-10$ and $+5$. So

$x^2 - 5x - 50 = (x - 10)(x + 5),$

with zeros $x = 10$ and $x = -5$.

$x^2 + 2x - 80$. Product $-80$, sum $2$: opposite signs, larger positive; product $80$, difference $2$ gives $10$ and $8$, so $+10$ and $-8$. So

$x^2 + 2x - 80 = (x + 10)(x - 8),$

with zeros $x = -10$ and $x = 8$.

Key takeaways

• The degree is the highest power of $x$; a value is a zero when substituting it gives $0$.
• A quadratic with zeros $\alpha, \beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta$, so the sum and product of the zeros are $-\tfrac{b}{a}$ and $\tfrac{c}{a}$.
• Many results follow from the sum and product alone, for example $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
• To factorise $x^2 + bx + c$, find two numbers whose product is $c$ and whose sum is $b$, using the four sign cases.