Can every cubic be written as three linear factors?

No: some cubics have only one linear factor, with the quadratic left over not factorising. So stop testing once you have found one factor, then deal with the quadratic separately rather than hunting for more roots.

GCSEFurther MathsAQAFlashcards2. Algebra & FunctionsPolynomials & Factor Theorem

Polynomials & Factor Theorem (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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FrontFactor Theorem
Define polynomial.

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Cards in this collection (18)

Define polynomial.
A polynomial is a sum of terms with non-negative integer powers of $x$ .
Its degree is the highest power present, so $5 x^{6} - 2 x^{2}$ has degree 6.
Why are $\frac{1}{x} + x^{3}$ and $x^{2} + 3 x + \sqrt{x}$ not polynomials?
Because $\frac{1}{x} = x^{- 1}$ has a negative power, and $\sqrt{x} = x^{\frac{1}{2}}$ has a non-integer power.
Every power in a polynomial has to be a whole number that is zero or more.
True or False?
The number 10 on its own is a polynomial.
True.
10 is a polynomial of degree 0, because it can be thought of as $10 x^{0}$ .
A polynomial only needs non-negative integer powers, and zero is one of them.
Complete the names for each degree of polynomial:
Degree 2: $______$
Degree 3: $______$
Degree 4: $______$
The completed names are:
Degree 2: quadratic
Degree 3: cubic
Degree 4: quartic
Degree 1 is linear, and degree 5 is quintic.
In $6 x^{4} - 3 x^{2} + 2 x + 8$ , what is the degree and what is the constant?
The degree is 4, since that is the highest power of $x$ present.
The constant is 8, the term with no $x$ in it, and the coefficient of $x^{2}$ is $- 3$ .
Expand $(x^{3} + 4 x) (x^{2} + 8 x + 3)$ .
Multiply every term in the first bracket by every term in the second, then collect like terms:
$x^{5} + 8 x^{4} + 7 x^{3} + 32 x^{2} + 12 x$
The $x^{3}$ terms are the only ones that combine, since $3 x^{3} + 4 x^{3} = 7 x^{3}$ .
Complete the Factor Theorem:
If $f (a) =______$ then $(x______a)$ is a factor of $f (x)$ .
The completed theorem is:
If $f (a) = 0$ then $(x - a)$ is a factor of $f (x)$ .
Watch the sign: $f (2) = 0$ gives the factor $(x - 2)$ , not $(x + 2)$ . It also works in reverse.
Show that $(x - 2)$ is a factor of $f (x) = x^{3} - 6 x^{2} + 11 x - 6$ .
Substitute $x = 2$ , which gives $f (2) = 8 - 24 + 22 - 6 = 0$ .
Because the result is zero, the Factor Theorem says $(x - 2)$ is a factor, with no factorising needed.
How do you test whether $(2 x - 3)$ is a factor of $f (x)$ ?
Set $2 x - 3 = 0$ to get $x = \frac{3}{2}$ , then work out $f (\frac{3}{2})$ .
If that comes to zero then $(2 x - 3)$ is a factor, by the Factor Theorem.
If $f (\frac{3}{2}) = 0$ , is $(2 x - 3)$ or $(3 x - 2)$ a factor?
The factor is $(2 x - 3)$ , because the denominator of the fraction becomes the coefficient of $x$ .
Substituting $\frac{b}{a}$ tests $(a x - b)$ , so $\frac{3}{2}$ tests $(2 x - 3)$ and never $(3 x - 2)$ .
$(2 x + 1)$ is a factor of $2 x^{2} + 7 x + 3$ . What is the other factor?
The other factor is $(x + 3)$ .
Only $x$ multiplied by $2 x$ gives the $2 x^{2}$ , and only $+ 3$ multiplied by $+ 1$ gives the $+ 3$ , so there is just one possibility.
Complete the structure used to factorise a cubic when one linear factor is known:
A cubic is a linear factor multiplied by a $______$ factor.
The completed statement is:
A cubic is a linear factor multiplied by a quadratic factor.
So write $f (x) \equiv (x - k) (a x^{2} + b x + c)$ and then find $a$ , $b$ and $c$ .
Given that $(x - 2)$ is a factor of $2 x^{3} + 7 x^{2} - 17 x - 10$ , factorise it fully.
Writing it as $(x - 2) (a x^{2} + b x + c)$ gives $a = 2$ and $c = 5$ by inspection, and equating the $x^{2}$ terms gives $b = 11$ .
The quadratic $2 x^{2} + 11 x + 5$ then factorises, so the answer is $(x - 2) (2 x + 1) (x + 5)$ .
Which values should you test when hunting for a linear factor of $x^{3} - 6 x^{2} - x + 30$ ?
Only the positive and negative whole numbers that divide 30, the constant term.
Working through them, $f (- 2) = 0$ , so $(x + 2)$ is a factor.
Why is hunting for a factor of $3 x^{3} + 4 x^{2} - 5 x - 2$ harder?
Because the 3 in front of $x^{3}$ means the factors need not all be of the form $(x \pm k)$ .
You may also have to try $(3 x + 1)$ , $(3 x - 1)$ , $(3 x + 2)$ and $(3 x - 2)$ .
True or False?
If $f (1)$ , $f (- 1)$ and $f (2)$ are all zero for a cubic, it factorises to $(x - 1) (x + 1) (x - 2)$ .
False.
For $f (x) = 2 x^{3} - 4 x^{2} - 2 x + 4$ all three are zero, but the cubic is $2 (x - 1) (x + 1) (x - 2)$ .
Those three brackets alone expand to a cubic starting with $x^{3}$ , so the leading coefficient still has to be accounted for.
Solve $3 x^{3} + 4 x^{2} - 5 x - 2 = 0$ .
$f (1) = 0$ , so $(x - 1)$ is a factor and the cubic factorises to $(x - 1) (x + 2) (3 x + 1)$ .
Setting each bracket equal to zero gives $x = 1$ , $x = - 2$ and $x = - \frac{1}{3}$ .
Can every cubic be written as three linear factors?
No: some cubics have only one linear factor, with the quadratic left over not factorising.
So stop testing once you have found one factor, then deal with the quadratic separately rather than hunting for more roots.

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