Polynomials and Quadratic Equations

Polynomials: Terms, Degree, Value and Zeros

A polynomial in $x$ is a sum of terms such as $5x^3$, $-2x$ and $7$, where every power of $x$ is a whole number (0, 1, 2, 3 and so on). So $4x^3 - 5x^2 + 7$ is a polynomial, but $x^2 + \frac{1}{x}$ and $3x + 2\sqrt{x}$ are not, because $\frac{1}{x} = x^{-1}$ and $\sqrt{x} = x^{\frac{1}{2}}$. The coefficients can be fractions or surds: $\sqrt{2}x^2 + \frac{1}{3}$ is still a polynomial. Only the powers of $x$ decide.

The words SSC uses. For $p(x) = 4x^3 - 5x^2 + 7$:

• the terms are $4x^3$, $-5x^2$ and $7$
• the coefficient of $x^2$ is $-5$ (keep the sign), and the coefficient of $x$ is 0 because there is no $x$ term
• the constant term is 7
• the degree is 3, the highest power of $x$, and the leading coefficient is 4

Degree	Name	Looks like
0	constant	$7$
1	linear	$3x - 2$
2	quadratic	$x^2 - 5x + 6$
3	cubic	$2x^3 + x - 1$
4	biquadratic	$x^4 - 16$

Degree of a product. When polynomials multiply, their degrees add. $(x^2 + 1)^3(2x - 5)^2$ has degree $2 \times 3 + 1 \times 2 = 8$. A sum has degree at most that of the bigger part, and it can drop when the top terms cancel: $(x^3 + 2x) + (5 - x^3) = 2x + 5$ has degree 1.

Two variables. Add the powers inside each term and take the largest. In $3x^2y^3 - 4xy + 9$ the first term has degree $2 + 3 = 5$, so the polynomial has degree 5.

Value at a point. Put the number in place of $x$:

$$p(2) = 4(8) - 5(4) + 7 = 19 \qquad p(-1) = -4 - 5 + 7 = -2$$

Three values come almost free:

• $p(0)$ is the constant term.
• $p(1)$ is the sum of all the coefficients. So the coefficients of $(3x - 2)^5$ add up to $(3 - 2)^5 = 1$, with no expanding.
• $p(-1)$ is the same sum with the signs of the odd powers flipped.

Zeros. A zero of $p(x)$ is a value of $x$ that makes $p(x) = 0$. The linear polynomial $ax + b$ has one zero, $-\frac{b}{a}$, so $3x - 2$ has the zero $\frac{2}{3}$. A polynomial of degree $n$ has at most $n$ zeros. To test a number, substitute it: for $q(x) = x^3 - 2x^2 - 5x + 6$, $q(1) = 1 - 2 - 5 + 6 = 0$, so 1 is a zero. In fact $q(x) = (x - 1)(x - 3)(x + 2)$, with zeros 1, 3 and $-2$.

Note that $(x^2 + 1)^3$ has degree 6, not 5: the power 2 is multiplied by 3, not added to it.

The Remainder Theorem

Divide $f(x)$ by $x - a$, and call the quotient $q(x)$ and the remainder $r$. Then

$$f(x) = (x - a)\,q(x) + r$$

Put $x = a$: the first part becomes zero, so $f(a) = r$. In words, the remainder when $f(x)$ is divided by $x - a$ is $f(a)$. No long division is needed.

Take $f(x) = x^3 - 4x^2 + 2x + 5$.

• Divided by $x - 3$: $f(3) = 27 - 36 + 6 + 5 = 2$.
• Divided by $x + 2$: $x + 2 = 0$ gives $x = -2$, so the remainder is $f(-2) = -8 - 16 - 4 + 5 = -23$.
• Divided by $2x - 1$: $2x - 1 = 0$ gives $x = \frac{1}{2}$, so the remainder is $f\left(\frac{1}{2}\right) = \frac{1}{8} - 1 + 1 + 5 = \frac{41}{8}$.

One rule covers every case: set the divisor equal to zero and put that value of $x$ into $f(x)$. For a divisor $ax - b$, use $x = \frac{b}{a}$.

Remainder given, find $k$. "When $x^3 + kx^2 - 3x + 4$ is divided by $x - 2$, the remainder is 10." Then $f(2) = 8 + 4k - 6 + 4 = 4k + 6 = 10$, so $k = 1$.

Two remainders, two unknowns. "$x^3 + ax^2 + bx + 6$ leaves remainder 4 when divided by $x - 1$ and remainder 12 when divided by $x + 1$." This gives $1 + a + b + 6 = 4$ and $-1 + a - b + 6 = 12$, that is $a + b = -3$ and $a - b = 7$. Adding the two, $2a = 4$, so $a = 2$ and $b = -5$.

High powers. The remainder when $x^{45} + 45$ is divided by $x + 1$ is $(-1)^{45} + 45 = -1 + 45 = 44$. An odd power of $-1$ is $-1$ and an even power is $+1$, so $x^{44} + 45$ would leave 46.

[Tier 2] Dividing by a quadratic. The remainder now has degree less than 2, so it looks like $px + q$. When the divisor splits, as $x^2 - 1 = (x - 1)(x + 1)$ does, put both values in. For $f(x) = x^{25} + 3$ divided by $x^2 - 1$: $f(1) = 4$ gives $p + q = 4$, and $f(-1) = 2$ gives $-p + q = 2$. So $q = 3$, $p = 1$, and the remainder is $x + 3$.

[SSC Tip] Dividing by $x + 3$ means putting $x = -3$, not $x = 3$. SSC nearly always keeps the value at $x = 3$ among the options.

The Factor Theorem: Finding an Unknown and Making a Polynomial Divisible

The factor theorem is the remainder theorem with remainder zero: $x - a$ is a factor of $f(x)$ exactly when $f(a) = 0$.

• Is $x - 2$ a factor of $x^3 - 3x^2 + 4$? $f(2) = 8 - 12 + 4 = 0$, so yes. Dividing out, $x^3 - 3x^2 + 4 = (x - 2)(x^2 - x - 2) = (x - 2)^2(x + 1)$.
• Find $k$ if $x + 3$ is a factor of $2x^3 + 5x^2 + kx - 6$. Put $x = -3$: $-54 + 45 - 3k - 6 = 0$, so $-15 - 3k = 0$ and $k = -5$.

Two factors given. "$x - 1$ and $x + 2$ are both factors of $x^3 + ax^2 + bx - 4$." From $f(1) = 0$: $a + b = 3$. From $f(-2) = 0$: $-8 + 4a - 2b - 4 = 0$, so $2a - b = 6$. Adding, $3a = 9$, so $a = 3$ and $b = 0$. Check: $(x - 1)(x + 2)^2 = x^3 + 3x^2 - 4$. When the question gives a quadratic factor such as $x^2 + x - 2$, split it into $(x - 1)(x + 2)$ first and use both values.

What must be subtracted or added. "What must be subtracted from $x^3 + 2x^2 - 5x + 9$ to make it exactly divisible by $x - 1$?" The remainder is $f(1) = 1 + 2 - 5 + 9 = 7$, so subtract 7. Taking away the remainder leaves an exact multiple. "What must be added?" asks for the same change with the other sign: add $-7$.

Quick factor tests.

• If the coefficients add up to 0, then $f(1) = 0$ and $x - 1$ is a factor.
• If the coefficients of the even powers (the constant counts as even) add up to the same as those of the odd powers, then $f(-1) = 0$ and $x + 1$ is a factor.

Factorising a cubic by trial. Try small divisors of the constant term: $\pm 1$, $\pm 2$, $\pm 3$ and so on. For $x^3 + 2x^2 - 11x - 12$, the even-power coefficients give $2 + (-12) = -10$ and the odd-power ones give $1 + (-11) = -10$. They match, so $x + 1$ is a factor, and

$$x^3 + 2x^2 - 11x - 12 = (x + 1)(x^2 + x - 12) = (x + 1)(x + 4)(x - 3)$$

To divide quickly, compare coefficients: $(x + 1)(x^2 + px + q)$ must give $x^3 + 2x^2 - 11x - 12$, so $q = -12$, and $1 + p = 2$ gives $p = 1$.

In "which of these is a factor" questions, test each option with the factor theorem, starting with the quick tests for $x - 1$ and $x + 1$. One substitution per option is faster than factorising the whole thing.

Using a Given Equation to Cut Down High Powers

Some questions give an equation in $x$ and ask for the value of a bigger expression. Solving for $x$ is usually slow or messy. Use the equation to bring the powers down instead.

Method 1: spot a perfect cube. Keep these two in mind:

$$(x - 1)^3 = x^3 - 3x^2 + 3x - 1 \qquad (x + 2)^3 = x^3 + 6x^2 + 12x + 8$$

"If $a^3 - 6a^2 + 12a - 3 = 0$, find $(a - 2)^6$." The first three terms are those of $(a - 2)^3 = a^3 - 6a^2 + 12a - 8$. So the equation is $(a - 2)^3 + 5 = 0$, which gives $(a - 2)^3 = -5$, and then $(a - 2)^6 = 25$.

"If $x^3 - 6x^2 + 12x = 16$, find the real value of $x$." Subtract 8 from both sides: $(x - 2)^3 = 8$, so $x - 2 = 2$ and $x = 4$.

What to look for: the coefficients of $x^3$, $x^2$ and $x$ are $1$, $3h$ and $3h^2$ (with matching signs), which are the first three terms of $(x + h)^3$.

Method 2: replace the square. "If $x^2 - 3x - 1 = 0$, find $x^3 - 2x^2 - 4x + 5$." From the equation, $x^2 = 3x + 1$. Then

$$x^3 = x \cdot x^2 = 3x^2 + x = 3(3x + 1) + x = 10x + 3$$

so $x^3 - 2x^2 - 4x + 5 = (10x + 3) - 2(3x + 1) - 4x + 5 = 6$. Every $x$ cancels, which is a sign you are on track.

Method 3: divide by the given expression. The same question by division:

$$x^3 - 2x^2 - 4x + 5 = (x^2 - 3x - 1)(x + 1) + 6$$

The bracket is 0, so the value is the remainder, 6. To find the quotient, match the top term first ($x^3$ needs $x$), then the next one ($-2x^2$ needs $+1$).

When $x$ is a surd. If $x = 2 + \sqrt{3}$, first build its equation, $x^2 - 4x + 1 = 0$ (the Reciprocal Forms (x + 1/x) and Surd Values section shows how), and then reduce: $x^3 - 4x^2 + x + 7 = x(x^2 - 4x + 1) + 7 = 7$. Questions such as $x^2 + 1 = x$ with powers like $x^{14}$ are also in that section.

Quadratic Equations: Solving by Factorising and by the Formula

A quadratic equation has the form $ax^2 + bx + c = 0$ with $a \neq 0$. It has at most two roots.

Splitting the middle term. Find two numbers whose product is $ac$ and whose sum is $b$. For $6x^2 - x - 15 = 0$, $ac = -90$ and $b = -1$, so the numbers are $-10$ and $9$:

$$6x^2 - 10x + 9x - 15 = 2x(3x - 5) + 3(3x - 5) = (2x + 3)(3x - 5)$$

Either bracket can be zero, so $x = -\frac{3}{2}$ or $x = \frac{5}{3}$.

The formula. When the two numbers do not come easily, use

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

For $2x^2 - 6x + 1 = 0$: $x = \frac{6 \pm \sqrt{36 - 8}}{4} = \frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm \sqrt{7}}{2}$.

Do not divide by $x$. $x^2 = 5x$ becomes $x(x - 5) = 0$, so $x = 0$ or $x = 5$. Dividing both sides by $x$ throws the root 0 away.

Equations that turn into quadratics.

• A square on one side: $(x - 2)^2 = 49$ gives $x - 2 = \pm 7$, so $x = 9$ or $x = -5$.
• Fractions: in $\frac{1}{x} - \frac{1}{x + 3} = \frac{1}{6}$ the left side is $\frac{3}{x(x + 3)}$, so $x^2 + 3x = 18$, which is $(x + 6)(x - 3) = 0$: $x = 3$ or $x = -6$.
• A square root: $\sqrt{x + 7} = x - 5$. Squaring gives $x + 7 = x^2 - 10x + 25$, so $x^2 - 11x + 18 = 0$ and $x = 2$ or $x = 9$. Now check both in the original: $x = 9$ gives $4 = 4$, but $x = 2$ gives $3 = -3$, which is false. Only $x = 9$ works. Squaring can create a false root, so always check.
• Powers of $x^2$: in $x^4 - 13x^2 + 36 = 0$ put $y = x^2$. Then $(y - 4)(y - 9) = 0$, so $x^2 = 4$ or $x^2 = 9$, and $x = \pm 2$ or $x = \pm 3$.

[SSC Tip] With options in front of you, substitute them instead of solving. Watch the signs: if the roots are $-\frac{3}{2}$ and $\frac{5}{3}$, an option with $\frac{3}{2}$ and $-\frac{5}{3}$ is the sign-slip trap.

Sum and Product of Roots, and Expressions Built from Them

For $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

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

Why: $a(x - \alpha)(x - \beta) = ax^2 - a(\alpha + \beta)x + a\alpha\beta$, and this must match $ax^2 + bx + c$. For $2x^2 - 7x + 4 = 0$ the sum is $\frac{7}{2}$ and the product is $\frac{4}{2} = 2$, with no solving at all.

Most questions ask for something built from $\alpha$ and $\beta$ that stays the same when you swap them. Rewrite it using only the sum and the product. These are the identities of the Square and Cube Identities section, with $a + b$ and $ab$ now coming from the coefficients.

Asked	Rewrite as
$\alpha^2 + \beta^2$	$(\alpha + \beta)^2 - 2\alpha\beta$
$\frac{1}{\alpha} + \frac{1}{\beta}$	$\frac{\alpha + \beta}{\alpha\beta}$
$(\alpha - \beta)^2$	$(\alpha + \beta)^2 - 4\alpha\beta$
$\alpha^3 + \beta^3$	$(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$	$\frac{\alpha^2 + \beta^2}{\alpha\beta}$
$\frac{1}{\alpha^2} + \frac{1}{\beta^2}$	$\frac{\alpha^2 + \beta^2}{(\alpha\beta)^2}$
$\alpha^2\beta + \alpha\beta^2$	$\alpha\beta(\alpha + \beta)$
$(\alpha + 1)(\beta + 1)$	$\alpha\beta + (\alpha + \beta) + 1$

With $\alpha + \beta = \frac{7}{2}$ and $\alpha\beta = 2$:

• $\alpha^2 + \beta^2 = \frac{49}{4} - 4 = \frac{33}{4}$
• $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{7}{2} \div 2 = \frac{7}{4}$
• $(\alpha - \beta)^2 = \frac{49}{4} - 8 = \frac{17}{4}$, so $\alpha - \beta = \pm\frac{\sqrt{17}}{2}$
• $\alpha^3 + \beta^3 = \frac{343}{8} - 3 \times 2 \times \frac{7}{2} = \frac{343}{8} - 21 = \frac{175}{8}$

[SSC Tip] Get the sign of the sum right: for $x^2 - 9x + 14 = 0$ the sum is $+9$, not $-9$. And $\alpha - \beta$ has two possible signs unless the question says which root is larger; with $\alpha > \beta$, take the positive square root.

Forming an Equation from Its Roots, and One Root Given

If the roots are known, the equation is

$$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$$

• Roots 4 and $-7$: sum $-3$, product $-28$, so $x^2 + 3x - 28 = 0$.
• Roots $\frac{2}{3}$ and $-\frac{1}{2}$: sum $\frac{1}{6}$, product $-\frac{1}{3}$. Multiply through by 6 to clear fractions: $6x^2 - x - 2 = 0$.
• One root $3 + \sqrt{2}$, with rational coefficients: the other root must be $3 - \sqrt{2}$. Sum 6, product $9 - 2 = 7$, so $x^2 - 6x + 7 = 0$.

New roots from old ones. Let $\alpha$, $\beta$ be the roots of $ax^2 + bx + c = 0$.

New roots	New equation
$\frac{1}{\alpha}$, $\frac{1}{\beta}$	$cx^2 + bx + a = 0$ (coefficients written backwards)
$-\alpha$, $-\beta$	$ax^2 - bx + c = 0$
$k\alpha$, $k\beta$	$ax^2 + kbx + k^2c = 0$
$\alpha + h$, $\beta + h$	put $x - h$ in place of $x$

For anything else, find the new sum and product. If $\alpha$, $\beta$ are the roots of $x^2 - 5x + 3 = 0$, then $\alpha^2$ and $\beta^2$ have sum $25 - 6 = 19$ and product $9$, so their equation is $x^2 - 19x + 9 = 0$.

One root given. Substitute the root to find the unknown, then use the sum or the product for the other root. "One root of $x^2 - kx + 15 = 0$ is 3." Then $9 - 3k + 15 = 0$, so $k = 8$, and since the product is 15 the other root is 5.

Special relations between roots. For $ax^2 + bx + c = 0$:

• roots reciprocal to each other (product 1): $c = a$
• roots equal in size but opposite in sign (sum 0): $b = 0$
• one root zero: $c = 0$
• one root twice the other: $2b^2 = 9ac$; in general, one root $k$ times the other: $kb^2 = (k + 1)^2ac$

For $x^2 - 9x + p = 0$ with one root twice the other: $2 \times 81 = 9p$, so $p = 18$, and the roots are 3 and 6.

Nature of Roots: The Discriminant

The number under the square root in the formula decides what kind of roots you get:

$$D = b^2 - 4ac$$

Value of $D$	Roots
$D > 0$	two different real roots (rational if $D$ is a perfect square and $a$, $b$, $c$ are rational)
$D = 0$	two equal real roots, each $-\frac{b}{2a}$
$D < 0$	no real roots

• $3x^2 - 5x + 2 = 0$: $D = 25 - 24 = 1$, a perfect square, so two different rational roots (1 and $\frac{2}{3}$).
• $x^2 - 4x - 1 = 0$: $D = 16 + 4 = 20$, so two different real roots, both irrational.
• $4x^2 - 12x + 9 = 0$: $D = 144 - 144 = 0$, so the roots are equal: $\frac{3}{2}$ and $\frac{3}{2}$.
• $2x^2 + 3x + 4 = 0$: $D = 9 - 32 < 0$, so there are no real roots.

Equal roots, find $k$. Set $D = 0$.

• $x^2 + kx + 25 = 0$: $k^2 - 100 = 0$, so $k = \pm 10$.
• $kx^2 - 12x + 4 = 0$: $144 - 16k = 0$, so $k = 9$. ($k = 0$ is never allowed here, since the equation would stop being a quadratic.)
• $(k + 1)x^2 - 2(k - 1)x + 1 = 0$: $4(k - 1)^2 - 4(k + 1) = 0$, which simplifies to $k^2 - 3k = 0$, so $k = 0$ or $k = 3$. Both work: they give $(x + 1)^2 = 0$ and $(2x - 1)^2 = 0$.

Real roots, or no real roots. "Real roots" means $D \ge 0$. For $x^2 + 4x + k = 0$: $16 - 4k \ge 0$, so $k \le 4$. For no real roots, $k > 4$.

Perfect squares. $4x^2 + kx + 9$ is a perfect square exactly when $D = 0$: $k^2 = 144$, so $k = \pm 12$, giving $(2x \pm 3)^2$.

Zeros and Coefficients of a Cubic

[Tier 2] The sum-and-product idea carries over to cubics. Tier 1 asks only the simplest part of it, the sum of the zeros; Tier 2 goes further. For $ax^3 + bx^2 + cx + d$ with zeros $\alpha$, $\beta$, $\gamma$:

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

The signs go minus, plus, minus.

Take $2x^3 - 5x^2 - 14x + 8$, whose zeros are 4, $\frac{1}{2}$ and $-2$. The formulas agree:

• sum: $4 + \frac{1}{2} - 2 = \frac{5}{2}$, and $-\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}$
• products in pairs: $2 - 1 - 8 = -7$, and $\frac{c}{a} = \frac{-14}{2} = -7$
• product: $4 \times \frac{1}{2} \times (-2) = -4$, and $-\frac{d}{a} = -\frac{8}{2} = -4$

Built expressions. Same idea as for quadratics:

$$\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$$

$$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}$$

For the cubic above these give $\frac{25}{4} + 14 = \frac{81}{4}$ and $\frac{-7}{-4} = \frac{7}{4}$.

Forming a cubic. With the zeros known, write $x^3 - (\text{sum})x^2 + (\text{pair sum})x - (\text{product})$. Zeros 1, 2 and 3 give $x^3 - 6x^2 + 11x - 6$.

Two zeros known, find the third. "Two zeros of $x^3 - 4x^2 - 7x + 10$ are 1 and $-2$." All three zeros add up to 4, so the third is $4 - 1 + 2 = 5$.

Zeros in AP. Write them as $m - d$, $m$ and $m + d$; they add up to $3m$. "The zeros of $x^3 - 12x^2 + 39x - k$ are in AP. Find $k$." Here $3m = 12$, so 4 is a zero: $64 - 192 + 156 - k = 0$, which gives $k = 28$. The zeros turn out to be 1, 4 and 7.

If the $x^2$ term is missing, as in $x^3 - 7x + 6$, the zeros add up to 0. And the middle relation has a plus sign: writing $-\frac{c}{a}$ for the pair sum is the slip behind the usual wrong option.

Number Questions That Lead to a Quadratic, and HCF and LCM

Number questions. Call the number $n$, turn the sentence into an equation, and solve.

• "Three times a number, added to its square, gives 108." So $n^2 + 3n = 108$, that is $n^2 + 3n - 108 = 0$, which factorises as $(n + 12)(n - 9) = 0$: $n = 9$ or $n = -12$. If the question says "positive" or "natural", only 9 is allowed.
• "Two consecutive positive odd numbers have product 323." You could solve $n(n + 2) = 323$, but this is faster: the two numbers sit either side of $\sqrt{323}$, which is just under 18, so they are 17 and 19. Check: $17 \times 19 = 323$.
• "The squares of two consecutive positive integers add up to 365." Then $n^2 + (n + 1)^2 = 365$, so $2n^2 + 2n - 364 = 0$, that is $n^2 + n - 182 = 0$. This gives $(n + 14)(n - 13) = 0$, and the integers are 13 and 14.
• "Two numbers have sum 15 and product 56." They are the roots of $x^2 - 15x + 56 = 0$, which are 7 and 8.

Testing the options is often the fastest of all: an option either fits every condition or it does not.

HCF and LCM of polynomials. Factorise each polynomial completely. Then

• HCF: take the common factors, each with its lowest power
• LCM: take every factor that appears, each with its highest power

$x^2 - 9 = (x - 3)(x + 3)$ and $x^2 + x - 6 = (x + 3)(x - 2)$, so the HCF is $x + 3$ and the LCM is $(x - 3)(x + 3)(x - 2)$.

$(x - 1)^2(x + 2)$ and $(x - 1)(x + 2)^3$ have HCF $(x - 1)(x + 2)$ and LCM $(x - 1)^2(x + 2)^3$.

Just as with numbers, $\text{HCF} \times \text{LCM}$ equals the product of the two polynomials (up to a constant factor). So if the HCF and both polynomials are known, the LCM is the product divided by the HCF.

Throw out roots that break the question: a negative number of people, a fraction where whole numbers are needed, or a "positive integer" that comes out as $-12$.

Solved Examples

Question 1

What is the degree of the polynomial $(x^3 - 2)^2(x^2 + x + 1)^3$? (a) 5 (b) 12 (c) 36 (d) 10

Answer. When you multiply polynomials, the degrees add. The first bracket $(x^3 - 2)^2$ has top term $x^6$, so its degree is $3 \times 2 = 6$. The second bracket $(x^2 + x + 1)^3$ has top term $x^6$ too, so its degree is $2 \times 3 = 6$. Multiplying them gives degree $6 + 6$ = (b) 12. Multiplying the two sixes instead of adding them gives 36.

Question 2

If $p(x) = 2x^3 - 3x^2 + 5x - 4$, what is the value of $p(2) - p(-1)$? (a) $-4$ (b) 18 (c) 10 (d) 24

Answer. Put each number in carefully. $p(2) = 16 - 12 + 10 - 4 = 10$. For $p(-1)$, remember $(-1)^3 = -1$ and $(-1)^2 = 1$: $p(-1) = -2 - 3 - 5 - 4 = -14$. So $p(2) - p(-1) = 10 - (-14)$ = (d) 24.

Question 3

What is the remainder when $x^3 - 5x^2 + 4x + 9$ is divided by $x + 2$? (a) $-27$ (b) 5 (c) $-11$ (d) 27

Answer. By the remainder theorem, set the divisor to zero: $x + 2 = 0$ gives $x = -2$. Now put $x = -2$ in: $(-8) - 5(4) + 4(-2) + 9 = -8 - 20 - 8 + 9$ = (a) $-27$. The option 5 is the value at $x = 2$, the usual sign slip with $x + 2$.

Question 4

What is the remainder when $4x^3 - 2x^2 + x - 5$ is divided by $2x - 3$? (a) $-\frac{49}{2}$ (b) $\frac{3}{2}$ (c) $\frac{11}{2}$ (d) 88

Answer. Set $2x - 3 = 0$, so $x = \frac{3}{2}$. Then $4 \times \frac{27}{8} = \frac{27}{2}$, $2 \times \frac{9}{4} = \frac{9}{2}$, and the remainder is $\frac{27}{2} - \frac{9}{2} + \frac{3}{2} - 5 = \frac{21}{2} - 5$ = (c) $\frac{11}{2}$.

Question 5

When $x^3 + kx^2 - 7x + 6$ is divided by $x - 3$, the remainder is 30. What is the value of $k$? (a) $\frac{10}{3}$ (b) 2 (c) $-\frac{8}{3}$ (d) $\frac{14}{3}$

Answer. Divisor $x - 3$, so the remainder is the value at $x = 3$: $27 + 9k - 21 + 6 = 9k + 12$. This has to equal 30, so $9k = 18$ and $k$ = (b) 2. Check: with $k = 2$, $27 + 18 - 21 + 6 = 30$.

Question 6

The polynomial $2x^3 + ax^2 + bx - 3$ leaves remainder $-2$ when divided by $x - 1$ and remainder 2 when divided by $x + 1$. Find $a$ and $b$. (a) $a = -4$, $b = 3$ (b) $a = 4$, $b = -3$ (c) $a = 3$, $b = 4$ (d) $a = 3$, $b = -4$

Answer. Use the remainder theorem twice. At $x = 1$: $2 + a + b - 3 = -2$, so $a + b = -1$. At $x = -1$: $-2 + a - b - 3 = 2$, so $a - b = 7$. Add the two: $2a = 6$, so $a = 3$, and then $b = -1 - 3 = -4$. The answer is (d) $a = 3$, $b = -4$. $a = -4$, $b = 3$ also gives $a + b = -1$, but fails the second condition; always check both.

Question 7

If $x - 2$ is a factor of $x^3 - 3x^2 + kx - 10$, what is the value of $k$? (a) 7 (b) $-15$ (c) $-7$ (d) 9

Answer. A factor means the remainder is zero, so put $x = 2$ and set it equal to 0: $8 - 12 + 2k - 10 = 0$, which is $2k - 14 = 0$. So $k$ = (a) 7. Check: $x^3 - 3x^2 + 7x - 10 = (x - 2)(x^2 - x + 5)$.

Question 8

If $x^2 + 2x - 3$ is a factor of $x^3 + ax^2 + bx + 6$, find $a$ and $b$. (a) $a = -7$, $b = 0$ (b) $a = 2$, $b = -3$ (c) $a = 0$, $b = -7$ (d) $a = 1$, $b = -8$

Answer. First split the quadratic factor: $x^2 + 2x - 3 = (x - 1)(x + 3)$. So both $x = 1$ and $x = -3$ must make the cubic zero. At $x = 1$: $1 + a + b + 6 = 0$, so $a + b = -7$. At $x = -3$: $-27 + 9a - 3b + 6 = 0$, so $9a - 3b = 21$, that is $3a - b = 7$. Adding $a + b = -7$ and $3a - b = 7$ gives $4a = 0$, so $a = 0$ and $b = -7$: (c) $a = 0$, $b = -7$. Check: $x^3 - 7x + 6 = (x - 1)(x + 3)(x - 2)$.

Question 9

What must be subtracted from $2x^3 - 5x^2 + 4x + 7$ so that the result is exactly divisible by $x - 2$? (a) $-11$ (b) 11 (c) $-37$ (d) 7

Answer. The remainder on dividing by $x - 2$ is the value at $x = 2$: $16 - 20 + 8 + 7 = 11$. If you take the remainder away, what is left divides exactly, so subtract (b) 11. $-11$ would be the answer if the question asked what must be added.

Question 10

Which of the following is a factor of $x^3 - 7x - 6$? (a) $x - 1$ (b) $x + 3$ (c) $x - 2$ (d) $x + 2$

Answer. Test each option with the factor theorem. For $x - 1$ put $x = 1$: $1 - 7 - 6 = -12$, not zero. For $x + 3$ put $x = -3$: $-27 + 21 - 6 = -12$, not zero. For $x - 2$ put $x = 2$: $8 - 14 - 6 = -12$, not zero. For $x + 2$ put $x = -2$: $-8 + 14 - 6 = 0$. So the factor is (d) $x + 2$. In fact $x^3 - 7x - 6 = (x + 1)(x + 2)(x - 3)$.

Question 11

[Tier 2] What is the remainder when $x^{23} + 2x + 5$ is divided by $x^2 - 1$? (a) $3x + 5$ (b) 8 (c) $5x + 3$ (d) 2

Answer. Dividing by a quadratic leaves a remainder like $px + q$. Since $x^2 - 1 = (x - 1)(x + 1)$, use $x = 1$ and $x = -1$. At $x = 1$: $1 + 2 + 5 = 8$, so $p + q = 8$. At $x = -1$: $-1 - 2 + 5 = 2$, so $-p + q = 2$. Adding, $2q = 10$, so $q = 5$ and $p = 3$. The remainder is (a) $3x + 5$.

Question 12

What is the remainder when $x^{41} + 41$ is divided by $x + 1$? (a) 42 (b) 0 (c) 40 (d) 41

Answer. Put $x = -1$. Since 41 is odd, $(-1)^{41} = -1$, so the remainder is $-1 + 41$ = (c) 40.