Factoring out the greatest common factor (GCF)

LVL: FREE

MODULE: Polynomials and Functions

βœ–οΈ 1. Identifying the GCF of numerical coefficients and variable terms

πŸ” Finding the GCF

  • The GCF is the largest number and lowest power of variables that divides all terms.
  • For coefficients: find the largest number that divides all coefficients evenly.
  • For variables: take the smallest exponent that appears in all terms.
  • If a variable is missing from any term, it's not part of the GCF.
  • Check each term separately, then combine the results.

Example: In 12x3+18x212x^3 + 18x^2, the GCF is 6x26x^2 (coefficients: GCF of 12 and 18 is 6; variables: smallest power is x2x^2).

πŸ’‘ GCF = biggest shared piece from ALL terms

1. Identifying the GCF of numerical coefficients and variable terms

Identifying the GCF of numerical coefficients and variable terms

The greatest common factor (GCF) of a polynomial is the largest expression that divides all terms without remainder. It consists of the GCF of numerical coefficients multiplied by the lowest power of each common variable.

Intuition: Find the "biggest piece" shared by all termsβ€”both numbers and variablesβ€”that can be extracted uniformly.

Core Rules:

  • For coefficients: Find the largest integer dividing all numerical parts (e.g., GCF of 12, 18, 30 is 6)
  • For variables: Take the smallest exponent present in all terms (e.g., in x3x^3, x5x^5, x2x^2, use x2x^2)
  • If a variable appears in only some terms, it is not part of the GCF
  • The GCF of 1 and any integer is always 1

Consequence: Correctly identifying the GCF ensures complete factorization and prevents leaving common factors inside parentheses.

Example: For 18x4y2+24x3y518x^4y^2 + 24x^3y^5, the GCF is 6x3y26x^3y^2 (coefficients: GCF of 18 and 24 is 6; variables: x3x^3 and y2y^2 are smallest powers).

Progress0 / 3
Level 2
EXEC: ALGORITHM

Find the greatest common factor of the terms 15x415x^4 and 25x225x^2.

Deep reasoning
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βœ–οΈ 2. Dividing polynomial terms by the GCF to factor the expression

βœ‚οΈ Pulling Out the GCF

  • Write the GCF outside parentheses as a multiplier.
  • Divide each original term by the GCF to get what stays inside.
  • Subtract exponents when dividing variables: x5Γ·x2=x3x^5 \div x^2 = x^3.
  • The number of terms inside the parentheses equals the number of original terms.
  • Always check: multiplying back should give the original expression.

Example: 15x4βˆ’10x2=5x2(3x2βˆ’2)15x^4 - 10x^2 = 5x^2(3x^2 - 2) because 15x4Γ·5x2=3x215x^4 \div 5x^2 = 3x^2 and 10x2Γ·5x2=210x^2 \div 5x^2 = 2.

πŸ’‘ GCF goes outside, leftovers go inside

2. Dividing polynomial terms by the GCF to factor the expression

Dividing polynomial terms by the GCF to factor the expression

Factoring extracts the GCF by dividing each term by it and placing the GCF outside parentheses. The quotient forms the remaining polynomial inside.

Intuition: Reverse the distributive propertyβ€”pull out the shared factor, leaving what remains after division.

Core Rules:

  • Divide every term of the polynomial by the identified GCF
  • Write the result as: GCF Γ—\times (sum of quotients)
  • Each quotient must be simplified completely (e.g., 12x54x2=3x3\frac{12x^5}{4x^2} = 3x^3)
  • The number of terms inside parentheses equals the original term count

Consequence: Proper division ensures the factored form is equivalent to the original expression and cannot be factored further by common factors.

Example: Factor 15x3βˆ’10x215x^3 - 10x^2. GCF is 5x25x^2. Dividing: 15x35x2=3x\frac{15x^3}{5x^2} = 3x and 10x25x2=2\frac{10x^2}{5x^2} = 2. Result: 5x2(3xβˆ’2)5x^2(3x - 2).

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Level 2
EXEC: ALGORITHM

A student is factoring the polynomial 20x4+15x320x^4 + 15x^3. They correctly identify the GCF as 5x35x^3. What is the expression that remains inside the parentheses after dividing each term by the GCF?

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βœ–οΈ 3. Factoring out negative GCFs and common binomial factors

βž– Negative and Binomial GCFs

  • Factor out a negative when the first term's coefficient is negative to make it positive.
  • Factoring out βˆ’1-1 flips all signs inside the parentheses.
  • A binomial can be a GCF if it appears in every term.
  • Treat the binomial as a single unit when factoring.

Example: βˆ’6y+12=βˆ’6(yβˆ’2)-6y + 12 = -6(y - 2) and 3(a+b)+5x(a+b)=(a+b)(3+5x)3(a+b) + 5x(a+b) = (a+b)(3 + 5x).

πŸ’‘ Negatives flip signs; binomials act like single variables

3. Factoring out negative GCFs and common binomial factors

Factoring out negative GCFs and common binomial factors

When the leading coefficient is negative or terms share a binomial expression, factor out βˆ’1-1 times the GCF or the entire binomial.

Intuition: Extracting a negative flips signs inside parentheses; binomials factor like single variables when repeated.

Core Rules:

  • Negative GCF: Factor out βˆ’1-1 multiplied by the positive GCF to make the leading term inside positive (e.g., βˆ’6x2+9x=βˆ’3x(2xβˆ’3)-6x^2 + 9x = -3x(2x - 3))
  • Sign changes: Factoring out a negative reverses all signs inside parentheses
  • Binomial factors: Treat repeated binomials as single units (e.g., 3(x+2)+5y(x+2)=(x+2)(3+5y)3(x+2) + 5y(x+2) = (x+2)(3 + 5y))
  • Verify the leading term inside parentheses matches the desired form

Consequence: Negative factoring standardizes expressions for further operations; binomial factoring reveals hidden common structure.

Example: Factor βˆ’4a2+12a-4a^2 + 12a. GCF is βˆ’4a-4a. Result: βˆ’4a(aβˆ’3)-4a(a - 3).

Progress0 / 3
Level 2
EXEC: ALGORITHM

Factor the expression by factoring out the negative greatest common factor: βˆ’3x2+12x-3x^2 + 12x.

Deep reasoning
Ultra

βœ–οΈ 4. Verifying factoring by redistributing

βœ… Checking Your Work

  • Distribute the GCF back through the parentheses to verify.
  • Multiply the GCF by each term inside using the distributive property.
  • If you get the original expression, the factoring is correct.
  • This catches errors in signs or exponent arithmetic.

Example: Check 4x2(2xβˆ’3)4x^2(2x - 3): distribute to get 4x2β‹…2xβˆ’4x2β‹…3=8x3βˆ’12x24x^2 \cdot 2x - 4x^2 \cdot 3 = 8x^3 - 12x^2 βœ“

πŸ’‘ Multiply back = original means you're right

4. Verifying factoring by redistributing

Verifying factoring by redistributing

Verification applies the distributive property to the factored form to confirm it reproduces the original polynomial exactly.

Intuition: Multiply back outβ€”if factoring is correct, you recover the starting expression term-by-term.

Core Rules:

  • Multiply the GCF by each term inside the parentheses separately
  • Combine like terms if any appear after distribution
  • Every term from the original must reappear with identical coefficients and exponents
  • If results differ, recheck GCF identification or division steps

Consequence: Verification catches errors in sign handling, exponent arithmetic, or incomplete factorization before proceeding to further algebraic steps.

Example: Verify 6x2(2xβˆ’5)6x^2(2x - 5). Distribute: 6x2β‹…2x=12x36x^2 \cdot 2x = 12x^3 and 6x2β‹…(βˆ’5)=βˆ’30x26x^2 \cdot (-5) = -30x^2. Result: 12x3βˆ’30x212x^3 - 30x^2 matches the original.

Progress0 / 3
Level 2
EXEC: ALGORITHM

A student factored 4x2βˆ’12x4x^2 - 12x as 4x(xβˆ’3)4x(x - 3). Which expression shows the correct verification step by redistributing?

Deep reasoning
Ultra
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βœ–οΈ 5. Applications: Simplifying formulas in physics to isolate shared constants

🌍 Physics Formula Shortcuts

  • Factoring isolates common constants like gg, mm, or Ο€\pi in formulas.
  • Makes calculations faster when the constant appears multiple times.
  • In energy sums, factor out gg or mm to simplify before plugging in values.
  • Reduces arithmetic errors by computing shared factors once.

Example: Total potential energy mgh1+mgh2=mg(h1+h2)mgh_1 + mgh_2 = mg(h_1 + h_2) β€” compute mgmg once, then add heights.

πŸ’‘ Factor constants out = calculate once, use everywhere

5. Applications: Simplifying formulas in physics to isolate shared constants

Applications: Simplifying formulas in physics to isolate shared constants

Factoring extracts repeated physical constants from sums, clarifying relationships and simplifying calculations in formulas.

Intuition: When multiple energy terms share a constant like gg or mm, factor it out to see the combined effect of variable quantities.

Core Rules:

  • Identify constants appearing in all terms (e.g., gravitational acceleration gg, mass mm)
  • Factor the constant outside parentheses, leaving variable-dependent expressions inside
  • Simplifies substitution: Compute the parenthetical sum once, then multiply by the constant
  • Common in energy sums: mgh1+mgh2=mg(h1+h2)mgh_1 + mgh_2 = mg(h_1 + h_2)

Consequence: Factoring reduces computational steps and reveals that total potential energy depends on gg times the sum of heights, not individual products.

Example: Total potential energy for two masses at heights 10 m and 15 m: mg(10)+mg(15)=mg(10+15)=25mgmg(10) + mg(15) = mg(10 + 15) = 25mg joules.

Progress0 / 3
Level 2
EXEC: ALGORITHM

A system has two objects with the exact same mass mm and gravitational acceleration gg. Their heights are aa and bb. The total potential energy is mga+mgbmga + mgb.

Factor out the shared constants to simplify this expression. Write the factored algebraic expression.

Deep reasoning
Ultra

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