Polynomials path · A05

Before this lesson: Remainder and factor theorems, Equations and quadratic tools

Your goal: Solve structured equations without losing or inventing roots.

Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.

Solving polynomial equations: the key idea

Before expanding, look for a common factor, a familiar identity or a repeated expression. A polynomial in x² can be treated as a quadratic in t=x², but over the reals the new variable must satisfy t≥0. A reciprocal polynomial may suggest t=x+1/x, after excluding x=0. Factoring splits an equation into cases; dividing by an expression can erase a case. A complete solution identifies all candidates and checks that each satisfies the original domain and equation.

A worked example

Solve x⁴−5x²+4=0 over the reals.

Set t=x²≥0. Then t²−5t+4=(t−1)(t−4)=0, giving t=1 or 4. Both are nonnegative. Thus x=±1 or ±2. Substitution verifies all four, and the equivalent factorisation shows there are no others.

Your turn: change one thing

Solve x⁴+x²−2=0 over the reals.

Try this on paper before opening the explanation.

Compare your reasoning

Let t=x². Then (t+2)(t−1)=0. The value t=−2 is impossible for a real square, so t=1 and x=±1.

Pause and check

A trap to avoid: Squaring without checking or cancelling a factor that may be zero.

Practise and adjust the level

Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.

Interactive practice loads here. You can also use the complete question set below.

Write a complete argument

Find all real x satisfying x⁴−2x²+1=0. Explain the multiplicities.

Planning hint

List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.

Read the full solution after your attempt

The left side is (x²−1)²=(x−1)²(x+1)². A square vanishes exactly when its base vanishes, so the real roots are 1 and −1. Each has multiplicity two. These two distinct roots account for four roots with multiplicity.

My proof notebook

Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.

All nine practice questions

Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.

1. Which substitution simplifies x⁴−6x²+5=0?

Foundation

  1. t=x⁴+x
  2. t=x²
  3. t=x+4
  4. t=1/x with no restriction
Hint

The powers are 4,2,0.

Answer and reasoning

t=x². Writing t=x² turns the equation into t²−6t+5=0.

2. Which rule solves AB=0 over the reals?

Foundation

  1. A=1 or B=1
  2. A+B=0
  3. A=0 or B=0
  4. A=B
Hint

A product vanishes when a factor vanishes.

Answer and reasoning

A=0 or B=0. The real numbers have no zero divisors, so at least one factor must be zero.

3. Find all real roots of x⁴−5x²+4=0.

Core

  1. 1,2 only
  2. −4,−1,1,4
  3. 0,1,4
  4. −2,−1,1,2
Hint

Factor as a quadratic in x².

Answer and reasoning

−2,−1,1,2. (x²−1)(x²−4)=0 gives x=±1,±2.

4. How many real roots does x⁴+x²−2=0 have?

Core

  1. 1
  2. 4
  3. 2
  4. 0
Hint

Remember x²≥0.

Answer and reasoning

2. The possibilities for x² are 1 and −2. Only 1 is allowed, producing two roots.

5. A substitution t=x+1/x is made for real nonzero x. Which t values are possible?

Stretch

  1. t≤−2 or t≥2
  2. Every real t
  3. Only t≥0
  4. −2<t<2
Hint

Solve x²−tx+1=0 and require real roots.

Answer and reasoning

t≤−2 or t≥2. The discriminant t²−4 must be nonnegative. This gives the two stated ranges.

6. Why check candidates after squaring an equation?

Stretch

  1. Polynomials cannot be squared
  2. Only positive candidates are ever valid
  3. Opposite signs can have equal squares
  4. Squaring always loses every root
Hint

For example, 1²=(−1)².

Answer and reasoning

Opposite signs can have equal squares. The squared equation may accept an input whose two original sides have opposite signs.

7. When a product of real factors is zero, what follows?

Foundation

  1. At least one factor is zero
  2. Every factor is zero
  3. Every factor is one
  4. The factors are equal
Hint

Use the zero-product property.

Answer and reasoning

At least one factor is zero. A product of nonzero real numbers cannot be zero.

8. Solve x⁴−5x²+4=0 over ℝ.

Core

  1. x=1,4 only
  2. x=−4,4 only
  3. No real roots
  4. x=−2,−1,1,2
Hint

Set t=x² and factor.

Answer and reasoning

x=−2,−1,1,2. (t−1)(t−4)=0, so x²=1 or 4.

9. When substituting t=x² for real x, what restriction is required?

Stretch

  1. t≠1
  2. t≥0
  3. t<0
  4. t is an integer
Hint

A real square is nonnegative.

Answer and reasoning

t≥0. Negative t-roots do not correspond to real x.

Choose your next step

Continue to Vieta’s formulas. If this felt difficult, return to a prerequisite above. Every lesson stays open.

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