Expanding and Simplifying Polynomials
Every question with a fully worked solution.
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Theory
To expand a product of polynomials, multiply every term of one bracket by every term of the other, then collect like terms. This Year 10A maths page covers longer brackets, three brackets, cubes and squares, finding a single coefficient quickly and equating coefficients.
The distributive law \(a(b + c) = ab + ac\) extends to any brackets: every term in the first bracket multiplies every term in the second. A bracket of \(2\) terms times a bracket of \(3\) terms gives \(6\) products before simplifying.
Multiplying powers adds the indices, so \(x^3 \times x^2 = x^5\). The degree of a product is the sum of the degrees.
With three or more brackets, multiply two of them first, then multiply the result by the next. A power such as \((x - 2)^3\) means three equal brackets.
An identity is true for every value of \(x\). If two polynomials are identical, their coefficients of each power of \(x\) are equal. This is equating coefficients.
How to expand a product of polynomials
- Order each bracket in descending powers and write in any missing powers mentally (\(8 + x^5 - 10x^2\) becomes \(x^5 - 10x^2 + 8\)).
- Multiply each term of the shorter bracket by the whole longer bracket, one line per term. Watch the signs.
- Collect like terms, lining up equal powers, and write the answer in descending powers.
- Check by substituting a small value such as \(x = 1\) into the question and the answer.
Shortcuts and problem types
- One coefficient only: list the pairs of terms whose powers add to the power you need, multiply, and add (Example 3).
- Unknown constants: expand with the letters, group by powers, then equate coefficients. Use the leftover equation as a check.
- Condition on coefficients: if the coefficient of \(x^2\) must equal the coefficient of \(x\), write each in terms of \(k\) and solve.
- Models: volume \(=\) length \(\times\) width \(\times\) height in terms of \(x\); expand, then substitute and check with the dimensions.
- Show that, hence: prove the identity by expanding, then substitute a convenient number such as \(x = 10\) or \(x = 100\).
(i) \((3x + 2)(x^2 - 5x + 4)\)
(ii) \((x + 2)(x - 3)(2x + 1)\)
(i) Multiply each term of the first bracket by the whole second bracket.
| \(3x(x^2 - 5x + 4)\) | \(=\) | \(3x^3 - 15x^2 + 12x\) |
| \(2(x^2 - 5x + 4)\) | \(=\) | \(2x^2 - 10x + 8\) |
| \(\text{sum}\) | \(=\) | \(\boldsymbol{3x^3 - 13x^2 + 2x + 8}\) |
(ii) Multiply two brackets first, then multiply the result by the third.
| \((x + 2)(x - 3)\) | \(=\) | \(x^2 - x - 6\) |
Now multiply this by \((2x + 1)\).
| \(\) | \(=\) | \((x^2 - x - 6)(2x + 1)\) |
| \(\) | \(=\) | \(2x^3 + x^2 - 2x^2 - x - 12x - 6\) |
| \(\) | \(=\) | \(\boldsymbol{2x^3 - x^2 - 13x - 6}\) |
Check with \(x = 1\): \((3)(-2)(3) = -18\) and \(2 - 1 - 13 - 6 = -18\).
(i) \((2x + 1)^3\)
(ii) \((x^2 - 2x + 3)^2\)
(i) Square first, then multiply by the bracket once more.
| \((2x + 1)^2\) | \(=\) | \(4x^2 + 4x + 1\) |
| \((2x + 1)^3\) | \(=\) | \((4x^2 + 4x + 1)(2x + 1)\) |
| \(\) | \(=\) | \(8x^3 + 4x^2 + 8x^2\) |
| \(\) | \(\) | \({} + 4x + 2x + 1\) |
| \(\) | \(=\) | \(\boldsymbol{8x^3 + 12x^2 + 6x + 1}\) |
The \(1, 3, 3, 1\) pattern gives the same: \((2x)^3 + 3(2x)^2(1) + 3(2x)(1)^2 + 1^3\).
(ii) Write it as a product of two equal brackets: \(3 \times 3 = 9\) products, then collect.
| \(x^2(x^2 - 2x + 3)\) | \(=\) | \(x^4 - 2x^3 + 3x^2\) |
| \(-2x(x^2 - 2x + 3)\) | \(=\) | \(-2x^3 + 4x^2 - 6x\) |
| \(3(x^2 - 2x + 3)\) | \(=\) | \(3x^2 - 6x + 9\) |
| \(\text{sum}\) | \(=\) | \(\boldsymbol{x^4 - 4x^3 + 10x^2 - 12x + 9}\) |
Squaring each term is not enough: \(x^4 + 4x^2 + 9\) misses all the cross terms.
(i) the degree and leading coefficient
(ii) the constant term
(iii) the coefficient of \(x^3\).
(i) Leading terms: \(2x^3 \times x^2 = 2x^5\). Degree \(\boldsymbol{5}\), leading coefficient \(\boldsymbol{2}\).
(ii) Constant terms: \((-4) \times 5 = \boldsymbol{-20}\).
(iii) Take one term from each bracket so that the powers add to \(3\).
| \(2x^3 \times 5\) | \(=\) | \(10x^3\) |
| \((-x^2) \times (-2x)\) | \(=\) | \(2x^3\) |
| \(3x \times x^2\) | \(=\) | \(3x^3\) |
| \(10 + 2 + 3\) | \(=\) | \(\boldsymbol{15}\) |
The \(-4\) would need an \(x^3\) term in the second bracket, and there is none. Check every term of the first bracket so no pair is missed.
(ii) Show that \((x + 2)(x - 2)(x^2 + 4) = x^4 - 16\), and hence evaluate \(12 \times 8 \times 104\) without a calculator.
(i) Expand the left side and group by powers of \(x\).
| \(\) | \(=\) | \(x^3 + bx^2 - 2x + ax^2 + abx - 2a\) |
| \(\) | \(=\) | \(x^3 + (a + b)x^2 + (ab - 2)x - 2a\) |
Match each coefficient with the right side, starting with the easiest.
| \(\text{constant: } -2a\) | \(=\) | \(-6 \;\Rightarrow\; \boldsymbol{a = 3}\) |
| \(x^2\text{: } a + b\) | \(=\) | \(5 \;\Rightarrow\; \boldsymbol{b = 2}\) |
| \(\text{check } x\text{: } ab - 2\) | \(=\) | \(6 - 2 = 4\) |
(ii) The first two brackets are a difference of two squares.
| \((x + 2)(x - 2)\) | \(=\) | \(x^2 - 4\) |
| \((x^2 - 4)(x^2 + 4)\) | \(=\) | \(x^4 - 16\) |
With \(x = 10\): \(12 \times 8 \times 104 = 10^4 - 16 = \boldsymbol{9984}\).
Common pitfalls
Frequently asked questions
How do you multiply two polynomials?
Multiply every term in the first polynomial by every term in the second, add the indices when multiplying powers of x, then collect like terms and write the result in descending powers.
How do you expand three brackets?
Multiply any two of the brackets first to get a quadratic, then multiply that quadratic by the third bracket and collect like terms.
How do you expand a cube like (x + 2) cubed?
Square the bracket first, then multiply by the bracket once more. Or use \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\): \((x + 2)^3 = x^3 + 6x^2 + 12x + 8\).
How do you find one coefficient without expanding everything?
Choose one term from each bracket so that their powers of x add to the power you want. Multiply each pair and add the results.
What does equating coefficients mean?
If two polynomials are equal for every value of x, the coefficients of each power of x must be the same on both sides. Setting them equal gives equations for any unknown constants.