Define dividend , divisor and quotient in an algebraic division.

The dividend is the polynomial being divided, the divisor is what you are dividing by, and the quotient is the answer. Anything left over at the end is the remainder .

When is the remainder theorem quicker than algebraic division?

Whenever the remainder is all that is wanted, since one substitution replaces the entire division. Algebraic division is still needed when the quotient matters, because the theorem tells you nothing about it.

How many real roots can a cubic equation have?

Always either one or three , counting a repeated root as often as it occurs. A cubic can therefore never have no real roots at all, unlike a quadratic.

True or False? A pair of simultaneous equations always has exactly one solution.

False. A linear pair normally has one, but a linear equation paired with a quadratic one usually has two . It can also have just one, when the line is a tangent to the curve, or none at all when the two never meet.

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Equations & Identities (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

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FrontAlgebraic Division
Define polynomial.

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Cards in this collection (27)

Define polynomial.
A polynomial is an algebraic expression with a finite number of terms and non-negative integer indices only.
Both $3 x + 5$ and $2 x^{2} y - 4 y + 6$ qualify, and so does a plain number such as $8$ , which is a polynomial with a single constant term.
True or False?
$5 x^{- 3}$ is a polynomial term.
False.
A polynomial may only contain non-negative integer indices, and $- 3$ is negative.
For the same reason $\sqrt{x}$ is not one either, since $\sqrt{x} = x^{\frac{1}{2}}$ has a fractional index.
Define dividend, divisor and quotient in an algebraic division.
The dividend is the polynomial being divided, the divisor is what you are dividing by, and the quotient is the answer.
Anything left over at the end is the remainder.
How do you begin dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $x + 7$ ?
Divide the highest power of the dividend by the highest power of the divisor: $x^{3} \div x = x^{2}$ , which becomes the first term of the quotient.
Multiply the whole divisor by $x^{2}$ , subtract, then repeat on what is left, working down through the powers until only a number remains.
Complete the identity linking a polynomial to its quotient and remainder on division by $(x - a)$ :
$f (x) = (x - a) \times______+______$
The completed identity is:
$f (x) = (x - a) \times Q (x) + R$
Dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $(x + 3)$ , for instance, gives $(x + 3) (x^{2} + 3 x - 18) + 40$ .
What tells you that the divisor is a factor of the dividend?
The division comes out with a remainder of zero.
Dividing $x^{3} + 6 x^{2} - 9 x - 14$ by $(x + 7)$ leaves nothing over, so $x^{3} + 6 x^{2} - 9 x - 14 = (x + 7) (x^{2} - x - 2)$ .
Complete the factor theorem for a polynomial $f (x)$ :
$f (a) =______exactly when (x -______) is a factor$
The completed statement is:
$f (a) = 0 exactly when (x - a) is a factor$
The theorem works in both directions, so a known root hands you a factor and a known factor hands you a root.
To test whether $(x + 1)$ is a factor of $g (x)$ , what do you work out?
Work out $g (- 1)$ , not $g (1)$ : written as $(x - a)$ the factor has $a = - 1$ .
For $g (x) = 2 x^{3} + 3 x^{2} - x + 5$ that gives $g (- 1) = 7$ , so $(x + 1)$ is not a factor.
True or False?
Any integer solution of $f (x) = 0$ must divide the constant term of $f (x)$ .
True.
Multiplying out $(x - a) Q (x)$ makes the constant term of $f (x)$ equal to $- a$ times the constant term of $Q (x)$ , so $a$ has to be a factor of it.
That is why the factors of the constant term are the values worth trying first when hunting for a root.
Given that $(2 x - 3)$ is a factor of $h (x) = 2 x^{3} - b x^{2} + 7 x - 6$ , how do you find $b$ ?
Set $h (\frac{3}{2}) = 0$ , since $2 x - 3 = 2 (x - \frac{3}{2})$ makes $\frac{3}{2}$ the root to use.
That gives $\frac{45}{4} - \frac{9}{4} b = 0$ , so $b = 5$ .
What is the remainder when $f (x)$ is divided by $(x - a)$ ?
The remainder is $f (a)$ , so a single substitution finds it.
Dividing $2 x^{4} - 2 x^{3} - x^{2} - 3 x + 1$ by $(x - 2)$ leaves $f (2) = 7$ , with no division done at all.
What value do you substitute when the divisor is $(2 x + 1)$ ?
Substitute $x = - \frac{1}{2}$ , the value that makes the divisor zero.
Writing $2 x + 1 = 2 (x + \frac{1}{2})$ shows the root to use is $- \frac{1}{2}$ , so the remainder is $f (- \frac{1}{2})$ .
When is the remainder theorem quicker than algebraic division?
Whenever the remainder is all that is wanted, since one substitution replaces the entire division.
Algebraic division is still needed when the quotient matters, because the theorem tells you nothing about it.
How many real roots can a cubic equation have?
Always either one or three, counting a repeated root as often as it occurs.
A cubic can therefore never have no real roots at all, unlike a quadratic.
True or False?
A cubic equation can have exactly two distinct real solutions.
True.
It happens when one of the roots is repeated, so the graph touches the $x$ -axis at that value instead of crossing it.
Counting repeats there are still three roots, but only two different numbers among them.
What is the first thing you need before you can solve a cubic equation?
You need one root to start from, either given in the question or found by trying small integers such as $1$ , $- 1$ , $2$ and $- 2$ .
That root gives a linear factor, which can be taken out to leave a quadratic you can solve by any usual method.
How do you find the quadratic factor of $2 x^{3} - x^{2} - 8 x + 4$ given the root $x = 2$ ?
Write $2 x^{3} - x^{2} - 8 x + 4 = (x - 2) (p x^{2} + q x + r)$ and compare coefficients.
The $x^{3}$ terms give $p = 2$ and the constants give $- 2 r = 4$ , so $r = - 2$ ; the $x^{2}$ terms then give $q = 3$ .
Complete the fully factorised form of a cubic with three real roots $α$ , $β$ and $γ$ :
$a x^{3} + b x^{2} + c x + d =______(x - α) (x - β) (x -______)$
The completed form is:
$a x^{3} + b x^{2} + c x + d = a (x - α) (x - β) (x - γ)$
The leading coefficient $a$ has to be there: without it the right-hand side would always give $x^{3}$ a coefficient of $1$ .
What does it mean if the quadratic factor of a cubic has no real roots?
The linear factor supplies the only real root of the cubic.
That quadratic's discriminant is negative, so the graph of the cubic crosses the $x$ -axis exactly once.
Define simultaneous equations.
Simultaneous equations are two or more equations that have to be satisfied by the same values of the unknowns at the same time.
Two unknowns need two equations, and if either equation contains $x^{2}$ , $y^{2}$ or $x y$ the pair is called quadratic rather than linear.
How do you eliminate $x$ from $3 x + 2 y = 11$ and $2 x - y = 5$ ?
Scale both equations so the $x$ terms match: multiply the first by $2$ and the second by $3$ , giving $6 x + 4 y = 22$ and $6 x - 3 y = 15$ .
Eliminating the $6 x$ leaves $7 y = 7$ , so $y = 1$ , and substituting back gives $x = 3$ .
Complete the elimination rule for simultaneous equations:
When the matching terms have the same sign you $______$ the equations, and when they have opposite signs you $______$ them.
The completed rule is:
When the matching terms have the same sign you subtract the equations, and when they have opposite signs you add them.
Either way the aim is to make that pair of terms cancel, leaving one equation in one unknown.
How does substitution solve two linear simultaneous equations?
Rearrange one equation into $y = \dots$ , then replace every $y$ in the other equation with that expression, kept in brackets.
From $2 x - y = 5$ you get $y = 2 x - 5$ , and $3 x + 2 (2 x - 5) = 11$ solves to give $x = 3$ .
When solving one linear and one quadratic equation, which do you substitute into which?
Rearrange the linear equation to $y = \dots$ or $x = \dots$ and substitute that into the quadratic one.
Never the other way round: putting a quadratic into a linear equation raises the power instead of reducing the number of unknowns.
How do you solve $x^{2} + y^{2} = 25$ and $y - 2 x = 5$ ?
Substitute $y = 2 x + 5$ into the first equation to get $x^{2} + {(2 x + 5)}^{2} = 25$ , which simplifies to $5 x^{2} + 20 x = 0$ .
That gives $x = 0$ or $x = - 4$ , and putting each back into the linear equation gives $y = 5$ and $y = - 3$ .
True or False?
A pair of simultaneous equations always has exactly one solution.
False.
A linear pair normally has one, but a linear equation paired with a quadratic one usually has two.
It can also have just one, when the line is a tangent to the curve, or none at all when the two never meet.
Why are the solutions of a linear and quadratic pair given as matched pairs of values?
Because each solution is a single point satisfying both equations at once, so an $x$ value only means anything alongside its own $y$ .
Listing all the $x$ values and then all the $y$ values separately would suggest every combination works, when only the matched ones do.

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