How many real roots can a cubic equation have?

Always either one or three , counting a repeated root each time it occurs. Because a root can repeat, the number of different solutions may be one, two or three.

IGCSEMathsEdexcelMaths BFlashcardsAlgebraSolving Cubic Equations

Solving Cubic Equations (Edexcel IGCSE Maths B): Flashcards

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How many real roots can a cubic equation have?

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Cards in this collection (7)

How many real roots can a cubic equation have?
Always either one or three, counting a repeated root each time it occurs.
Because a root can repeat, the number of different solutions may be one, two or three.
A cubic has three real roots $α$ , $β$ and $γ$ . Fill in the two missing factors.
$a x^{3} + b x^{2} + c x + d = (x - α) (______) (______)$
The completed factorisation is:
$a x^{3} + b x^{2} + c x + d = (x - α) (x - β) (x - γ)$
Each real root gives one linear factor, so three real roots give three of them.
A cubic has exactly one real root. What does that tell you about its quadratic factor?
The quadratic factor has no real roots, so its discriminant $q^{2} - 4 p r$ is negative.
The cubic factorises as a linear factor multiplied by that quadratic, and the linear factor supplies the only real solution.
True or False?
The real roots of a cubic equation are the $x$ -intercepts of the graph of the cubic.
True.
A root is a value of $x$ that makes the expression equal zero, and the graph meets the $x$ -axis exactly where its value is zero.
So a cubic with only one real root crosses the $x$ -axis just once.
A cubic equation gives you no root to start from. How can you find one?
Substitute small integers such as $1$ , $- 1$ , $2$ and $- 2$ into the cubic until it comes out as zero.
A value that gives zero is a root, and by the factor theorem it hands you a factor to divide by.
You have found that $x = 2$ solves a cubic equation. What can you do next?
Since $x = 2$ is a root, $(x - 2)$ is a factor, so take it out to leave a quadratic factor.
Solving that quadratic gives any remaining solutions of the cubic.
To factorise $2 x^{3} - x^{2} - 8 x + 4$ you write it as $(x - 2) (p x^{2} + q x + r)$ . How do you find $p$ and $r$ quickly?
Compare the $x^{3}$ terms and the constant terms, since each involves only one unknown: $2 x^{3} = p x^{3}$ gives $p = 2$ , and $4 = - 2 r$ gives $r = - 2$ .
The value of $q$ then follows from comparing either the $x^{2}$ terms or the $x$ terms.

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