How does a polynomial division start?

Divide the highest power term of the polynomial by the highest power term of the divisor. That first result is the leading term of the answer, and the process is the long division, or bus stop, method used with numbers.

What do you repeat at each stage of a polynomial division?

Multiply the whole divisor by the term you have just found, then subtract that from what is above it. Bring down the next term, and divide again by the highest power of the divisor.

How do you expand two brackets that each contain several terms?

Multiply every term in the first bracket by every term in the second, then collect like terms. FOIL is just the special case of this for two brackets of exactly two terms, so it is a shortcut rather than a separate method.

What does it mean if a polynomial division leaves a remainder?

It means the divisor is not a factor of the polynomial. A division that finishes with zero left over shows the divisor divides exactly, so it is one of the polynomial's factors.

How do you expand a product of three brackets?

Expand one pair of brackets first and simplify that result. Then multiply what you get by the remaining bracket, again multiplying every term by every term.

Why can a remainder be linear when the divisor is a quadratic, but not when the divisor is linear?

Because you stop dividing once what is left has a lower degree than the divisor . Against a linear divisor that leaves only a constant, but against a quadratic divisor anything of degree 1 or 0 is already too small to divide.

How many real solutions can a cubic equation have?

One, two or three. It can never have none, and it can never have more than three, because the cubic factorises into at most three linear factors.

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Factors of Polynomials (Cambridge (CIE) IGCSE Additional Maths): Flashcards

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FrontOperations with Polynomials
Define a polynomial.

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Cards in this collection (26)

Define a polynomial.
A polynomial is an algebraic expression made of a finite number of terms, in which every index is a non-negative integer.
So $3 x + 5$ and $2 x^{2} y - 4 y + 6$ are polynomials, but anything carrying a fractional or negative index is not.
How does a polynomial division start?
Divide the highest power term of the polynomial by the highest power term of the divisor.
That first result is the leading term of the answer, and the process is the long division, or bus stop, method used with numbers.
True or False?
A plain number such as 8 is a polynomial.
True.
A single constant is a very simple polynomial: it has a finite number of terms and no index that breaks the rule.
You can think of it as $8 x^{0}$ , and 0 is a non-negative integer.
You are dividing $x^{4} + 11 x^{2} - 1$ by $x + 3$ . Why should you write it as $x^{4} + 0 x^{3} + 11 x^{2} + 0 x - 1$ ?
The zero terms hold the empty columns open, so that like terms stay lined up as you subtract.
Without them the powers slide out of alignment and terms get subtracted from the wrong things.
Why are $5 x^{- 3}$ and $\frac{3}{x}$ not polynomials?
Both carry a negative index, and a polynomial allows only non-negative integer indices.
The second one hides it, because $\frac{3}{x}$ is the same as $3 x^{- 1}$ .
What do you repeat at each stage of a polynomial division?
Multiply the whole divisor by the term you have just found, then subtract that from what is above it.
Bring down the next term, and divide again by the highest power of the divisor.
How do you expand two brackets that each contain several terms?
Multiply every term in the first bracket by every term in the second, then collect like terms.
FOIL is just the special case of this for two brackets of exactly two terms, so it is a shortcut rather than a separate method.
What does it mean if a polynomial division leaves a remainder?
It means the divisor is not a factor of the polynomial.
A division that finishes with zero left over shows the divisor divides exactly, so it is one of the polynomial's factors.
How do you expand a product of three brackets?
Expand one pair of brackets first and simplify that result.
Then multiply what you get by the remaining bracket, again multiplying every term by every term.
True or False?
When you divide a polynomial by a quadratic, the remainder must be a constant.
False.
The remainder can be linear, of the form $r x + s$ , as when $x^{4} + 4 x^{3} - x + 1$ is divided by $x^{2} - 2 x$ and leaves $23 x + 1$ .
It may still turn out to be a constant, which is just the case where $r = 0$ .
Complete the expansion:
$(x + 2) (x^{2} - 3 x + 4) = x^{3} -______-______+ 8$
The completed expansion is:
$(x + 2) (x^{2} - 3 x + 4) = x^{3} - x^{2} - 2 x + 8$
The $x^{2}$ terms come from $x \times (- 3 x)$ and $2 \times x^{2}$ , which combine to give $- 3 x^{2} + 2 x^{2} = - x^{2}$ .
Why can a remainder be linear when the divisor is a quadratic, but not when the divisor is linear?
Because you stop dividing once what is left has a lower degree than the divisor.
Against a linear divisor that leaves only a constant, but against a quadratic divisor anything of degree 1 or 0 is already too small to divide.
State the factor theorem, in both directions.
For a polynomial $f (x)$ : if $f (p) = 0$ then $(x - p)$ is a factor, and if $(x - p)$ is a factor then $f (p) = 0$ .
Both directions matter, because one lets you test a suspected factor and the other lets you find one.
How do you show that $(x - 2)$ is a factor of $f (x) = x^{3} + 6 x^{2} - 9 x - 14$ ?
Substitute $x = 2$ and show the result is zero:
$f (2) = 8 + 24 - 18 - 14 = 0$
By the factor theorem, $f (2) = 0$ is enough to conclude that $(x - 2)$ is a factor.
What is the connection between the factors of a polynomial and its roots?
Each linear factor gives a root, found by setting that factor equal to zero.
So $4 x^{3} + 8 x^{2} - 9 x - 18 = (x + 2) (2 x + 3) (2 x - 3)$ has roots $x = - 2$ , $x = - \frac{3}{2}$ and $x = \frac{3}{2}$ .
True or False?
If $f (3) = 0$ , then $(x + 3)$ is a factor of $f (x)$ .
False.
The factor is $(x - 3)$ , because that is the bracket which equals zero when $x = 3$ .
The sign in the bracket is always the opposite of the value you substituted.
State the remainder theorem.
When a polynomial $f (x)$ is divided by $(x - a)$ , the remainder is $f (a)$ .
Written in full, $f (x) = (x - a) Q (x) + f (a)$ , where $Q (x)$ is the quotient and $(x - a)$ is the divisor.
Find the remainder when $x^{2} - 2 x$ is divided by $(x - 3)$ .
Substitute $x = 3$ rather than carrying out the division:
$3^{2} - 2 \times 3 = 3$
The remainder theorem gives the remainder from a single substitution, so the whole division can be skipped when the remainder is all you need.
Why is the factor theorem a special case of the remainder theorem?
Because the remainder on dividing by $(x - a)$ is $f (a)$ , and a factor is exactly a divisor that leaves no remainder.
So when $f (a) = 0$ the statement $f (x) = (x - a) Q (x) + f (a)$ collapses to $f (x) = (x - a) Q (x)$ .
What is the remainder when $f (x)$ is divided by $(2 x - 1)$ ?
It is $f (\frac{1}{2})$ .
You always evaluate the polynomial at the value of $x$ that makes the divisor zero, and for $2 x - 1$ that value is $\frac{1}{2}$ rather than a whole number.
What is the strategy for solving a cubic equation?
Use the factor theorem to find one value $p$ with $f (p) = 0$ , which gives the factor $(x - p)$ .
Divide the cubic by that factor to leave a quadratic, then solve the quadratic in whichever way suits it.
The solutions are the values making each factor zero.
You know $(x - 2)$ is a factor. Complete the factorisation by inspection:
$x^{3} - 10 x^{2} + 12 x + 8 = (x - 2) (x^{2} -______x -______)$
The completed factorisation is:
$x^{3} - 10 x^{2} + 12 x + 8 = (x - 2) (x^{2} - 8 x - 4)$
Two coefficients are forced immediately: the $x^{3}$ term can only come from $x \times x^{2}$ , and the constant 8 can only come from $- 2 \times - 4$ .
After dividing a cubic by a linear factor you are left with a quadratic that will not factorise. What do you do?
Solve it with the quadratic formula, or by completing the square.
That gives the other two solutions, often in surd form, as with $x^{2} - 8 x - 4 = 0$ giving $x = 4 \pm 2 \sqrt{5}$ .
How many real solutions can a cubic equation have?
One, two or three.
It can never have none, and it can never have more than three, because the cubic factorises into at most three linear factors.
True or False?
The equation $(x - 2)^{2} (x + 1) = 0$ has three different solutions.
False.
There are only two different solutions, $x = 2$ and $x = - 1$ .
The factor $(x - 2)$ appears twice, so $x = 2$ is a repeated root rather than two separate ones.
A cubic equation has no constant term, such as $x^{3} + 4 x^{2} - 11 x = 0$ . What can you do straight away?
Take out a factor of $x$ , since a missing constant term means $x$ itself is a factor and $x = 0$ is a solution.
That leaves a quadratic to solve in the usual way, with no values to test at all.

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