Polynomials are the building blocks of most expressions in this course. Being quick and accurate at adding, subtracting, and multiplying them makes everything that follows easier, from simplifying rational expressions to working with quadratic functions.
A polynomial is a sum of terms like 3x2, −5x, and 2. Each term has a coefficient (the number) and a variable part.
Like terms have exactly the same variable part: 3x2 and −7x2 are like terms, but 3x2 and 3x are not. You can only add or subtract like terms.
The degree of a polynomial is its highest exponent. 4x3−x+6 has degree 3.
- Adding: remove the brackets and combine like terms.
- Subtracting: the minus sign applies to every term in the second bracket. Change the sign of each term, then combine like terms.
(5x−2)−(3x−4)=5x−2−3x+4=2x+2
Multiply every term in the first bracket by every term in the second, then combine like terms. When you multiply powers with the same base, add the exponents: x⋅x2=x3.
(x+3)(x−5)=x2−5x+3x−15=x2−2x−15
These patterns come up constantly, so they’re worth memorizing:
(a+b)2(a−b)2(a+b)(a−b)=a2+2ab+b2=a2−2ab+b2=a2−b2
The middle term 2ab is the one students forget: (x+3)2 is not x2+9.
Two expressions are equivalent if they’re equal for every value of the variable. A quick test: substitute a number (not 0 or 1, which can hide mistakes) into both. If the results differ, they aren’t equivalent. If they match, that’s good evidence, and simplifying both fully confirms it.
Let P=3x2−5x+2 and Q=x2+7x−9. Find P+Q and P−Q.
Solution.
P+Q=3x2−5x+2+x2+7x−9=4x2+2x−7
P−Q=3x2−5x+2−x2−7x+9=2x2−12x+11change every sign in Q
Expand and simplify (2x−3)(x+4) and (3x−5)2.
Solution.
(2x−3)(x+4)=2x2+8x−3x−12=2x2+5x−12
For the square, use (a−b)2=a2−2ab+b2 with a=3x and b=5:
(3x−5)2=9x2−2(3x)(5)+25=9x2−30x+25
Expand and simplify (x+2)(x2−3x+5).
Solution. Multiply each term of x+2 by all three terms of the trinomial:
(x+2)(x2−3x+5)=x3−3x2+5x+2x2−6x+10=x3−x2−x+10
Simplify 2(x−3)2−(x+1)(x−1).
Solution. Expand each part first, keeping the second product in brackets so the minus sign reaches every term:
2(x−3)2−(x+1)(x−1)=2(x2−6x+9)−(x2−1)=2x2−12x+18−x2+1=x2−12x+19
Check with x=1: the original gives 2(4)−(2)(0)=8, and the answer gives 1−12+19=8. ✓
Subtracting only the first term. (5x−2)−(3x−4) is 5x−2−3x+4. The minus sign changes every sign in the bracket.
Squaring a binomial term by term. (x+3)2=x2+6x+9, not x2+9. Write it as (x+3)(x+3) if you’re unsure.
Combining unlike terms. x2+x can’t be simplified. It is not x3 or 2x2.
Adding exponents when you should keep them. When multiplying, add exponents: x2⋅x3=x5. When adding like terms, keep them: x2+x2=2x2, not x4.
Dropping a negative. In (2x−3)(x+4), the product of −3 and 4 is −12. Keep track of the sign on every term.
1. (Warm-up) Simplify (4x2+3x−1)+(2x2−5x+6).
Solution
6x2−2x+5
2. (Warm-up) Simplify (5a2−2a+3)−(a2+4a−7).
Solution
5a2−2a+3−a2−4a+7=4a2−6a+10
3. (Warm-up) Expand −3x(2x2−x+4).
Solution
−6x3+3x2−12x
4. (Core) Expand and simplify (3x+2)(2x−5).
Solution
6x2−15x+4x−10=6x2−11x−10
5. (Core) Expand (2y−7)2.
Solution
(2y)2−2(2y)(7)+72=4y2−28y+49
6. (Core) Expand and simplify (x−3)(x2+2x−4).
Solution
(x−3)(x2+2x−4)=x3+2x2−4x−3x2−6x+12=x3−x2−10x+12
7. (Core) Simplify 3(x+2)2−2(x−1)(x+4).
Solution
3(x+2)2−2(x−1)(x+4)=3(x2+4x+4)−2(x2+3x−4)=3x2+12x+12−2x2−6x+8=x2+6x+20
8. (Core) Are (x+3)2 and x2+9 equivalent? Explain.
Solution
No. Test x=1: (1+3)2=16, but 12+9=10.
Expanding shows why: (x+3)2=x2+6x+9, which has an extra 6x.
9. (Challenge) A rectangle is (2x+3) m long and (x+1) m wide. A square with sides of x m is cut out of one corner. Write a simplified expression for the area that’s left.
Solution
(2x+3)(x+1)−x2=2x2+2x+3x+3−x2=x2+5x+3The remaining area is (x2+5x+3) m².
10. (Challenge) Expand and simplify (x+1)3.
Solution
Write it as (x+1)2(x+1):
(x+1)3=(x2+2x+1)(x+1)=x3+x2+2x2+2x+x+1=x3+3x2+3x+1