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Family Table Math

Adding, Subtracting, and Multiplying Polynomials

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 3x23x^2, −5x-5x, and 22. Each term has a coefficient (the number) and a variable part.

Like terms have exactly the same variable part: 3x23x^2 and −7x2-7x^2 are like terms, but 3x23x^2 and 3x3x are not. You can only add or subtract like terms.

The degree of a polynomial is its highest exponent. 4x3−x+64x^3 - x + 6 has degree 33.

  • 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(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=x3x \cdot x^2 = x^3.

(x+3)(x−5)=x2−5x+3x−15=x2−2x−15(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15

These patterns come up constantly, so they’re worth memorizing:

(a+b)2=a2+2ab+b2(a−b)2=a2−2ab+b2(a+b)(a−b)=a2−b2\begin{aligned} (a + b)^2 &= a^2 + 2ab + b^2 \\ (a - b)^2 &= a^2 - 2ab + b^2 \\ (a + b)(a - b) &= a^2 - b^2 \end{aligned}

The middle term 2ab2ab is the one students forget: (x+3)2(x + 3)^2 is not x2+9x^2 + 9.

Two expressions are equivalent if they’re equal for every value of the variable. A quick test: substitute a number (not 00 or 11, 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+2P = 3x^2 - 5x + 2 and Q=x2+7x−9Q = x^2 + 7x - 9. Find P+QP + Q and P−QP - Q.

Solution.

P+Q=3x2−5x+2+x2+7x−9=4x2+2x−7P + Q = 3x^2 - 5x + 2 + x^2 + 7x - 9 = 4x^2 + 2x - 7 P−Q=3x2−5x+2−x2−7x+9change every sign in Q=2x2−12x+11\begin{aligned} P - Q &= 3x^2 - 5x + 2 - x^2 - 7x + 9 && \text{change every sign in } Q \\ &= 2x^2 - 12x + 11 \end{aligned}

Expand and simplify (2x−3)(x+4)(2x - 3)(x + 4) and (3x−5)2(3x - 5)^2.

Solution.

(2x−3)(x+4)=2x2+8x−3x−12=2x2+5x−12(2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12

For the square, use (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 with a=3xa = 3x and b=5b = 5:

(3x−5)2=9x2−2(3x)(5)+25=9x2−30x+25(3x - 5)^2 = 9x^2 - 2(3x)(5) + 25 = 9x^2 - 30x + 25

Expand and simplify (x+2)(x2−3x+5)(x + 2)(x^2 - 3x + 5).

Solution. Multiply each term of x+2x + 2 by all three terms of the trinomial:

(x+2)(x2−3x+5)=x3−3x2+5x+2x2−6x+10=x3−x2−x+10\begin{aligned} (x + 2)(x^2 - 3x + 5) &= x^3 - 3x^2 + 5x + 2x^2 - 6x + 10 \\ &= x^3 - x^2 - x + 10 \end{aligned}

Simplify 2(x−3)2−(x+1)(x−1)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\begin{aligned} 2(x - 3)^2 - (x + 1)(x - 1) &= 2(x^2 - 6x + 9) - (x^2 - 1) \\ &= 2x^2 - 12x + 18 - x^2 + 1 \\ &= x^2 - 12x + 19 \end{aligned}

Check with x=1x = 1: the original gives 2(4)−(2)(0)=82(4) - (2)(0) = 8, and the answer gives 1−12+19=81 - 12 + 19 = 8. ✓

Subtracting only the first term. (5x−2)−(3x−4)(5x - 2) - (3x - 4) is 5x−2−3x+45x - 2 - 3x + 4. The minus sign changes every sign in the bracket.

Squaring a binomial term by term. (x+3)2=x2+6x+9(x + 3)^2 = x^2 + 6x + 9, not x2+9x^2 + 9. Write it as (x+3)(x+3)(x + 3)(x + 3) if you’re unsure.

Combining unlike terms. x2+xx^2 + x can’t be simplified. It is not x3x^3 or 2x22x^2.

Adding exponents when you should keep them. When multiplying, add exponents: x2⋅x3=x5x^2 \cdot x^3 = x^5. When adding like terms, keep them: x2+x2=2x2x^2 + x^2 = 2x^2, not x4x^4.

Dropping a negative. In (2x−3)(x+4)(2x - 3)(x + 4), the product of −3-3 and 44 is −12-12. Keep track of the sign on every term.

1. (Warm-up) Simplify (4x2+3x−1)+(2x2−5x+6)(4x^2 + 3x - 1) + (2x^2 - 5x + 6).

Solution6x2−2x+56x^2 - 2x + 5

2. (Warm-up) Simplify (5a2−2a+3)−(a2+4a−7)(5a^2 - 2a + 3) - (a^2 + 4a - 7).

Solution5a2−2a+3−a2−4a+7=4a2−6a+105a^2 - 2a + 3 - a^2 - 4a + 7 = 4a^2 - 6a + 10

3. (Warm-up) Expand −3x(2x2−x+4)-3x(2x^2 - x + 4).

Solution−6x3+3x2−12x-6x^3 + 3x^2 - 12x

4. (Core) Expand and simplify (3x+2)(2x−5)(3x + 2)(2x - 5).

Solution6x2−15x+4x−10=6x2−11x−106x^2 - 15x + 4x - 10 = 6x^2 - 11x - 10

5. (Core) Expand (2y−7)2(2y - 7)^2.

Solution(2y)2−2(2y)(7)+72=4y2−28y+49(2y)^2 - 2(2y)(7) + 7^2 = 4y^2 - 28y + 49

6. (Core) Expand and simplify (x−3)(x2+2x−4)(x - 3)(x^2 + 2x - 4).

Solution(x−3)(x2+2x−4)=x3+2x2−4x−3x2−6x+12=x3−x2−10x+12\begin{aligned} (x - 3)(x^2 + 2x - 4) &= x^3 + 2x^2 - 4x - 3x^2 - 6x + 12 \\ &= x^3 - x^2 - 10x + 12 \end{aligned}

7. (Core) Simplify 3(x+2)2−2(x−1)(x+4)3(x + 2)^2 - 2(x - 1)(x + 4).

Solution3(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\begin{aligned} 3(x + 2)^2 - 2(x - 1)(x + 4) &= 3(x^2 + 4x + 4) - 2(x^2 + 3x - 4) \\ &= 3x^2 + 12x + 12 - 2x^2 - 6x + 8 \\ &= x^2 + 6x + 20 \end{aligned}

8. (Core) Are (x+3)2(x + 3)^2 and x2+9x^2 + 9 equivalent? Explain.

Solution

No. Test x=1x = 1: (1+3)2=16(1 + 3)^2 = 16, but 12+9=101^2 + 9 = 10.

Expanding shows why: (x+3)2=x2+6x+9(x + 3)^2 = x^2 + 6x + 9, which has an extra 6x6x.

9. (Challenge) A rectangle is (2x+3)(2x + 3) m long and (x+1)(x + 1) m wide. A square with sides of xx 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+3\begin{aligned} (2x + 3)(x + 1) - x^2 &= 2x^2 + 2x + 3x + 3 - x^2 \\ &= x^2 + 5x + 3 \end{aligned}

The remaining area is (x2+5x+3)(x^2 + 5x + 3) m².

10. (Challenge) Expand and simplify (x+1)3(x + 1)^3.

Solution

Write it as (x+1)2(x+1)(x + 1)^2(x + 1):

(x+1)3=(x2+2x+1)(x+1)=x3+x2+2x2+2x+x+1=x3+3x2+3x+1\begin{aligned} (x + 1)^3 &= (x^2 + 2x + 1)(x + 1) \\ &= x^3 + x^2 + 2x^2 + 2x + x + 1 \\ &= x^3 + 3x^2 + 3x + 1 \end{aligned}