From Products of Brackets to Standard Form
Polynomials appear throughout the MPM2D course — in quadratic relations, area problems, and algebraic modelling. Before you can graph or factor a quadratic, you need to be comfortable writing it in expanded form. This lesson builds directly on the distributive property and combining like terms that you practised in Grade 9 (MTH1W). You will extend those ideas to multiply two binomials together and to recognise patterns that speed up your work. By the end, you will be able to take any product of two first-degree brackets and rewrite it as a neat second-degree polynomial in standard form.
What you will learn
- Identify the parts of a polynomial expression and recall key vocabulary.
- Expand a product of two binomials using the distributive property.
- Expand special products, including a binomial squared and a difference of squares.
- Collect like terms to write a simplified second-degree polynomial in standard form.
Vocabulary and Prerequisites Review
A polynomial is an expression made of terms, where each term is a number, a variable, or a product of numbers and variables with whole-number exponents. Examples:
3x2−5x+7 and
2x−4.
The degree of a term is the exponent on its variable. A second-degree polynomial (also called a quadratic polynomial) has a highest-degree term of degree 2, such as
4x2 or
−x2. The standard form of a quadratic polynomial is
ax2+bx+c, where
a,
b, and
c are real numbers and
a=0.
A monomial has one term: for example,
5x or
−3x2. A binomial has exactly two terms: for example,
x+4 or
2x−1. Expanding means removing brackets by multiplying; simplifying means collecting like terms so no two terms share the same variable part.
The key tool you will use is the distributive property:
a(b+c)=ab+ac. In Grade 9 you used it to expand a monomial times a binomial. Here you will use it twice in a row to expand a binomial times a binomial.
ax^2 + bx + c
- A second-degree (quadratic) polynomial has degree 2 as its highest power.
- Standard form is ax2+bx+c with a=0.
- Monomial: one term. Binomial: two terms.
- The distributive property a(b+c)=ab+ac is the engine of every expansion.
Expanding a Product of Two Binomials
When you multiply two binomials, treat the first binomial as a single object being distributed over the second. For example, to expand
(x+3)(x+5), rewrite it as
x(x+5)+3(x+5). Now apply the distributive property to each group separately.
Carrying out each multiplication:
x⋅x=x2,
x⋅5=5x,
3⋅x=3x,
3⋅5=15. This gives
x2+5x+3x+15. Finally, collect the like terms
5x and
3x to get the simplified result
x2+8x+15.
A helpful memory device is to track the four individual multiplications in order: First terms, Outer terms, Inner terms, Last terms. Many students call this FOIL. It works only for binomial times binomial, but the underlying reason it works is simply the distributive property applied twice.
Notice the result is always a second-degree polynomial when both binomials are first-degree. That is because multiplying
x⋅x always produces
x2, raising the degree by one.
(a + b)(c + d) = ac + ad + bc + bd
- Distribute the entire first binomial over each term in the second binomial.
- Four partial products are created; collect the two middle terms if they are like terms.
- The product of two first-degree binomials is always a second-degree polynomial.
- FOIL is a memory device, not a new rule — it rests on the distributive property.
Special Products Worth Recognising
Two patterns appear so often in MPM2D that recognising them saves time and reduces errors. The first is the perfect-square binomial. When you square a binomial such as
(x+4)2, you are multiplying
(x+4)(x+4). Expanding gives
x2+4x+4x+16=x2+8x+16. In general,
(a+b)2=a2+2ab+b2 and
(a−b)2=a2−2ab+b2.
The second pattern is the difference of squares. When one binomial is a sum and the other is the matching difference — such as
(x+5)(x−5) — the two middle terms cancel:
x2−5x+5x−25=x2−25. In general,
(a+b)(a−b)=a2−b2. The result has no
x term at all.
It is important to verify these patterns by expanding fully at least once rather than just memorising the end results. Understanding why the middle terms disappear in the difference of squares (they are equal in size but opposite in sign) helps you avoid errors.
One common trap:
(x+4)2 is NOT
x2+16. The middle term
2(x)(4)=8x is always present. Squaring a binomial always produces three terms unless the context causes cancellation.
- Perfect-square binomial: (a+b)2=a2+2ab+b2.
- Perfect-square binomial (subtraction): (a−b)2=a2−2ab+b2.
- Difference of squares: (a+b)(a−b)=a2−b2.
- Always verify a pattern by expanding fully the first time you use it.
Collecting Like Terms and Writing Standard Form
After expanding, you often have four or more terms. Collecting like terms means grouping terms with identical variable parts and adding their coefficients. Only terms with the exact same variable and exponent are like terms:
3x2 and
−x2 are like terms;
3x2 and
3x are not.
Once like terms are collected, write the polynomial in standard form by arranging terms from highest degree to lowest: the
x2 term first, then the
x term, then the constant. For example,
15+3x−5x+x2 becomes
x2−2x+15 in standard form.
Sometimes you need to expand an expression that contains a coefficient in front of the brackets, or even a subtraction between two expanded products. Always deal with any coefficient or sign outside a bracket before combining results. For instance, in
3(x+1)(x−2)−4, first expand
(x+1)(x−2), then multiply every term by
3, then subtract
4.
- Like terms share the same variable and the same exponent.
- Standard form: ax2+bx+c, highest degree first.
- Apply any outer coefficient after expanding the brackets.
- Handle subtraction between groups carefully — distribute the negative sign.
Putting It All Together: A Strategy for Any Expansion
When you face a more complex expression, a clear sequence of steps prevents errors. Step 1 — expand each pair of brackets using the distributive property, writing out all four partial products. Step 2 — apply any coefficient or negative sign sitting outside the brackets to every term inside the expanded group. Step 3 — collect like terms. Step 4 — write the result in standard form and double-check by substituting a simple number (such as
x=1) into both the original expression and your answer to confirm they match.
The substitution check at the end is especially useful. If the original expression gives
6 at
x=1 and your simplified version also gives
6 at
x=1, you have strong evidence your algebra is correct. If the values differ, you know to look for an error before moving on.
Keeping a neat layout — one operation per line, terms lined up in columns — makes it much easier to spot missing terms or sign errors. Many mistakes in polynomial expansion come from rushing the middle step and losing a term or flipping a sign.
- Expand → Apply outer coefficients → Collect like terms → Standard form.
- A substitution check (e.g. x=1) is a quick way to catch errors.
- Write one operation per line to keep track of every term.
- Sign errors are the most common source of mistakes — treat subtraction carefully.
Summary of Expansion Patterns
| Pattern Name | Bracket Form | Expanded Result | Key Feature |
|---|
| General binomial product | (a+b)(c+d) | ac+ad+bc+bd | Four partial products; collect like terms |
| Perfect-square (sum) | (a+b)2 | a2+2ab+b2 | Middle term is always 2ab |
| Perfect-square (difference) | (a−b)2 | a2−2ab+b2 | Middle term is −2ab; constant is positive |
| Difference of squares | (a+b)(a−b) | a2−b2 | Middle terms cancel; no x term remains |
Worked example
Expanding and Simplifying a Product of Two Binomials
Expand and simplify
(2x−3)(x+7), then write the result in standard form.
- Distribute the first term of the first binomial
Take
2x and multiply it by each term in the second binomial
(x+7).
- Distribute the second term of the first binomial
Take
−3 (keep the negative sign attached) and multiply it by each term in the second binomial.
- Write all four partial products together
Combine the results of the two distributions into one unsimplified expression.
- Collect like terms
The terms
14x and
−3x share the same variable part
x, so add their coefficients:
14+(−3)=11.
- State the result in standard form
The expression is already arranged from highest to lowest degree, so the simplified standard form is shown below.
Answer: 2x2+11x−21 Check: Substitute
x=1 into the original:
(2(1)−3)(1+7)=(−1)(8)=−8. Substitute
x=1 into the answer:
2(1)2+11(1)−21=2+11−21=−8. Both sides match, so the expansion is correct.
Worked example
Expanding an Expression with an Outer Coefficient and a Binomial Squared
Expand and simplify
3(x−4)2−2(x+1)(x−1), then write in standard form.
- Expand the perfect-square binomial
Write
(x−4)2 as
(x−4)(x−4) and expand:
x⋅x=x2,
x⋅(−4)=−4x,
(−4)⋅x=−4x,
(−4)(−4)=16. Collecting the two middle terms gives the expansion of the squared binomial.
- Apply the outer coefficient 3
Multiply every term of
x2−8x+16 by
3.
- Expand the difference of squares
(x+1)(x−1) is a difference of squares pattern with
a=x and
b=1. The middle terms cancel:
x2+x−x−1=x2−1.
- Apply the outer coefficient −2
Multiply every term of
x2−1 by
−2. Be careful with the sign:
−2×(−1)=+2.
- Combine both expanded groups
Write the two expanded groups side by side, ready to collect like terms.
- Collect like terms and write in standard form
Combine the
x2 terms:
3x2−2x2=x2. The
x term is
−24x with no matching term. Combine the constants:
48+2=50.
Answer: x2−24x+50 Check: Substitute
x=0 into the original:
3(0−4)2−2(0+1)(0−1)=3(16)−2(1)(−1)=48+2=50. Substitute
x=0 into the answer:
(0)2−24(0)+50=50. Both match, confirming the result.
Common mistakes and how to avoid them
Writing
(x+5)2=x2+25 and forgetting the middle term.
Correction: Always expand the square as
(x+5)(x+5) to get
x2+10x+25. The middle term
2(x)(5)=10x is never zero unless the binomial itself is zero.
Dropping the negative sign when distributing, for example writing
(x−3)(x+2)=x2+2x−3x−6 correctly but then writing
x2−x+6 instead of
x2−x−6.
Correction: Check each constant multiplication carefully. Here
(−3)(+2)=−6, not
+6.
Forgetting to multiply an outer coefficient across all terms after expanding brackets, for example writing
2(x+3)(x−1)=2x2−x+3x−3=2x2+2x−3 instead of correctly obtaining
2x2+4x−6.
Correction: First expand the brackets to get
x2+2x−3, then multiply every single term by
2:
2x2+4x−6.
Treating
3x2 and
3x as like terms and adding them to get
6x2 or
6x.
Correction: Like terms must share the exact same variable and exponent.
3x2 and
3x have different exponents and cannot be combined.
Writing the final answer out of standard form, for example
15−2x+x2.
Correction: Rearrange so that the highest-degree term comes first:
x2−2x+15.
Lesson summary
- A second-degree polynomial has highest degree 2 and is written in standard form as ax2+bx+c.
- Expanding a binomial product uses the distributive property twice, producing four partial products.
- Three special patterns — perfect-square sum, perfect-square difference, and difference of squares — follow predictable results that you can verify by full expansion.
- After expanding, collect only true like terms (same variable, same exponent) before writing the final answer.
- Always apply an outer coefficient or negative sign to every term of the expanded group.
- A substitution check (replacing x with a simple number in both the original and simplified expressions) is a reliable way to catch errors.
Check your understanding
Question 1
Which expression is the fully expanded and simplified form of
(x+6)(x−2)?
- x2+4x−12
- x2−4x−12
- x2+4x+12
- x2−8x−12
Show answer and explanation
x2+4x−12 Expanding:
x⋅x+x(−2)+6⋅x+6(−2)=x2−2x+6x−12=x2+4x−12. The outer and inner terms are
−2x+6x=+4x, and the constant is
−12.
Question 2
What is the simplified form of
(3x−1)2?
- 9x2+1
- 9x2−6x+1
- 9x2+6x+1
- 3x2−6x+1
Show answer and explanation
9x2−6x+1 Expanding
(3x−1)(3x−1):
(3x)2=9x2; middle term
=2(3x)(−1)=−6x; last term
=(−1)2=1. Result:
9x2−6x+1.
Question 3
Expand and simplify
2(x+3)(x−3).
- 2x2−18
- 2x2+18
- 2x2−6x−18
- 2x2−12x−18
Show answer and explanation
(x+3)(x−3) is a difference of squares:
x2−9. Multiply by
2:
2x2−18. There is no
x term because the middle terms cancelled.
Question 4
Which expression is equivalent to
(x+4)(x+2)−3(x+1)?
- x2+3x+5
- x2+3x+8
- x2+6x+5
- x2+3x−5
Show answer and explanation
x2+3x+5 First expand
(x+4)(x+2)=x2+2x+4x+8=x2+6x+8. Then expand
3(x+1)=3x+3. Subtract:
x2+6x+8−3x−3=x2+3x+5.
Key terms
- Polynomial
- An expression made up of terms where each term is a product of a number and a variable raised to a whole-number exponent.
- Degree
- The highest exponent on the variable in a polynomial. A second-degree polynomial has a highest exponent of 2.
- Standard form (quadratic)
- The arrangement ax2+bx+c where terms are written from highest to lowest degree.
- Binomial
- A polynomial with exactly two terms, such as x−5 or 3x+2.
- Distributive property
- The rule a(b+c)=ab+ac, which allows a factor outside brackets to be multiplied into each term inside.
- Like terms
- Terms that have exactly the same variable raised to exactly the same exponent. Only like terms can be combined by adding or subtracting their coefficients.
- Perfect-square binomial
- The square of a binomial, such as (a+b)2, which expands to a2+2ab+b2.
- Difference of squares
- The product (a+b)(a−b), which simplifies to a2−b2 because the middle terms cancel.
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About this lesson and its review
Published by DoAssignment. This reviewed lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic Q9. It is a study resource, not an official curriculum publication.
Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.
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