Factoring quadratic trinomials

LVL: FREE

MODULE: Polynomials and Functions

[EXEC: MICRO_CORE]

✖️ 1. Factoring monic trinomials using the sum-product method

🎯 Factoring Monic Trinomials

  • A monic trinomial has the form x2+bx+cx^2 + bx + c where the x2x^2 coefficient is 1.
  • Find two numbers that multiply to cc and add to bb.
  • Write the factored form as (x+m)(x+n)(x + m)(x + n) where mm and nn are those two numbers.
  • If both numbers are positive, both signs in factors are positive.
  • If cc is negative, the two numbers have opposite signs.

Example: Factor x2+5x+6x^2 + 5x + 6. Numbers 2 and 3 multiply to 6 and add to 5, so (x+2)(x+3)(x + 2)(x + 3).

💡 Think: "What two numbers multiply to the last term and add to the middle coefficient?"

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1. Factoring monic trinomials using the sum-product method

Factoring Monic Trinomials Using the Sum-Product Method

A monic trinomial has the form x2+bx+cx^2 + bx + c where the leading coefficient is 1. The sum-product method factors this expression by finding two integers whose product equals cc and whose sum equals bb.

The goal is to rewrite x2+bx+cx^2 + bx + c as (x+m)(x+n)(x + m)(x + n) where mm and nn satisfy specific conditions.

Core Rules:

  • Find integers mm and nn such that m⋅n=cm \cdot n = c (product condition)
  • Verify that m+n=bm + n = b (sum condition)
  • Write the factorization as (x+m)(x+n)(x + m)(x + n)
  • If no such integers exist, the trinomial is irreducible over integers

This method works because expanding (x+m)(x+n)(x + m)(x + n) yields x2+(m+n)x+mnx^2 + (m+n)x + mn, matching the original form when m+n=bm+n = b and mn=cmn = c.

Example: Factor x2+7x+12x^2 + 7x + 12. We need m⋅n=12m \cdot n = 12 and m+n=7m + n = 7. The pair m=3,n=4m=3, n=4 works: 3⋅4=123 \cdot 4 = 12 and 3+4=73+4=7. Thus x2+7x+12=(x+3)(x+4)x^2 + 7x + 12 = (x+3)(x+4).

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LVL_2
EXEC: ALGORITHM

Factor the monic trinomial x2+8x+15x^2 + 8x + 15 using the sum-product method.

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✖️ 2. Factoring non-monic trinomials using the ac-method or grouping

🔧 Factoring Non-Monic Trinomials

  • A non-monic trinomial has the form ax2+bx+cax^2 + bx + c where a≠1a \neq 1.
  • Use the ac-method: multiply aa and cc to get the target product.
  • Find two numbers that multiply to acac and add to bb.
  • Rewrite the middle term using those two numbers, then factor by grouping.
  • Group the first two terms and last two terms separately.
  • Factor out the common binomial from both groups.

Example: Factor 2x2+7x+32x^2 + 7x + 3. Here ac=6ac = 6 and b=7b = 7. Numbers 6 and 1 work. Rewrite as 2x2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

💡 Remember: "Multiply aa times cc first, then split the middle!"

[EXEC: DEEP_COMPUTE]

2. Factoring non-monic trinomials using the ac-method or grouping

Factoring Non-Monic Trinomials Using the AC-Method or Grouping

A non-monic trinomial has the form ax2+bx+cax^2 + bx + c where a≠1a \neq 1. The ac-method converts this into a four-term expression that can be factored by grouping.

The strategy involves splitting the middle term bxbx into two terms whose coefficients multiply to acac and add to bb.

Core Rules:

  • Compute the product acac (multiply the leading and constant coefficients)
  • Find integers mm and nn where m⋅n=acm \cdot n = ac and m+n=bm + n = b
  • Rewrite ax2+bx+cax^2 + bx + c as ax2+mx+nx+cax^2 + mx + nx + c
  • Factor by grouping: extract common factors from the first two terms and last two terms separately

Grouping reveals a common binomial factor, yielding the final factorization.

Example: Factor 2x2+7x+32x^2 + 7x + 3. Here ac=2⋅3=6ac = 2 \cdot 3 = 6 and b=7b=7. The pair m=6,n=1m=6, n=1 satisfies 6⋅1=66 \cdot 1=6 and 6+1=76+1=7. Rewrite: 2x2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)2x^2 + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3).

TASK_1[0 / 3]
LVL_2
EXEC: ALGORITHM

For the trinomial 3x2+14x+83x^2 + 14x + 8, the ac-method requires splitting the middle term into two terms. What are the two integers mm and nn that should be used?

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✖️ 3. Recognizing and factoring perfect square trinomials

✨ Perfect Square Trinomials

  • A perfect square trinomial has the form a2+2ab+b2a^2 + 2ab + b^2 or a2−2ab+b2a^2 - 2ab + b^2.
  • It factors as (a+b)2(a + b)^2 or (a−b)2(a - b)^2 respectively.
  • Check: the first and last terms must be perfect squares.
  • The middle term must equal twice the product of the square roots.
  • If the middle term is positive, use (a+b)2(a + b)^2; if negative, use (a−b)2(a - b)^2.

Example: x2+6x+9x^2 + 6x + 9 has square roots xx and 3, and 2(x)(3)=6x2(x)(3) = 6x, so it factors as (x+3)2(x + 3)^2.

💡 Visual cue: "First square, last square, middle is twice the product!"

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3. Recognizing and factoring perfect square trinomials

Recognizing and Factoring Perfect Square Trinomials

A perfect square trinomial is a quadratic expression that equals the square of a binomial: (px+q)2=p2x2+2pqx+q2(px + q)^2 = p^2x^2 + 2pqx + q^2. Recognition requires verifying that the first and last terms are perfect squares and the middle term equals twice their product.

This special structure allows immediate factorization without trial methods.

Core Rules:

  • Verify the first term ax2ax^2 is a perfect square: a=p2a = p^2 for some integer or rational pp
  • Verify the constant term cc is a perfect square: c=q2c = q^2 for some integer or rational qq
  • Check if the middle coefficient satisfies b=2pqb = 2pq (or b=−2pqb = -2pq for negative cases)
  • Factor as (px+q)2(px + q)^2 if b=2pqb = 2pq, or (px−q)2(px - q)^2 if b=−2pqb = -2pq

Perfect square trinomials arise frequently in completing the square and geometric applications.

Example: Factor 9x2+12x+49x^2 + 12x + 4. Here 9x2=(3x)29x^2 = (3x)^2, 4=224 = 2^2, and 12=2⋅3⋅212 = 2 \cdot 3 \cdot 2. Thus 9x2+12x+4=(3x+2)29x^2 + 12x + 4 = (3x + 2)^2.

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RSN: PATTERNEXEC: ALGORITHM

Factor the perfect square trinomial: x2+10x+25x^2 + 10x + 25.

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✖️ 4. Identifying irreducible (prime) quadratics over integers

🚫 Irreducible Quadratics

  • A quadratic is irreducible (or prime) if it cannot be factored using integers.
  • Check the discriminant Δ=b2−4ac\Delta = b^2 - 4ac.
  • If Δ\Delta is not a perfect square, the quadratic is irreducible over integers.
  • If Δ<0\Delta < 0, there are no real roots at all.
  • Always try factoring first; if no integer pairs work, it is likely irreducible.

Example: x2+x+1x^2 + x + 1 has Δ=1−4=−3\Delta = 1 - 4 = -3, which is negative, so it cannot be factored over integers.

💡 Quick test: "No perfect square discriminant means no integer factors!"

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4. Identifying irreducible (prime) quadratics over integers

Identifying Irreducible (Prime) Quadratics Over Integers

A quadratic trinomial ax2+bx+cax^2 + bx + c is irreducible over integers (or prime) if it cannot be expressed as a product of two non-constant polynomials with integer coefficients. Not all quadratics factor into integer binomials.

The discriminant Δ=b2−4ac\Delta = b^2 - 4ac provides a criterion for factorability over integers.

Core Rules:

  • Compute the discriminant Δ=b2−4ac\Delta = b^2 - 4ac
  • If Δ\Delta is not a perfect square, the quadratic is irreducible over integers
  • If Δ\Delta is a perfect square, the quadratic may factor (verify using sum-product or ac-method)
  • A negative discriminant always indicates irreducibility over integers (and over reals)

Irreducible quadratics still have real or complex roots via the quadratic formula, but these roots are irrational or non-real.

Example: Consider x2+x+1x^2 + x + 1. The discriminant is Δ=12−4(1)(1)=−3\Delta = 1^2 - 4(1)(1) = -3, which is negative. Therefore x2+x+1x^2 + x + 1 is irreducible over integers.

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LVL_2
EXEC: ALGORITHM

Based on the discriminant test, which of the following quadratics is irreducible over integers?

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✖️ 5. Applications: Finding equilibrium points where supply-demand quadratics meet

📊 Supply-Demand Equilibrium

  • Equilibrium occurs where supply equals demand, meaning S(q)=D(q)S(q) = D(q).
  • Set the two quadratic expressions equal and move all terms to one side.
  • Factor the resulting quadratic to find the equilibrium quantity.
  • Only positive solutions make sense in real-world contexts.
  • Substitute the quantity back into either equation to find the equilibrium price.

Example: If supply is q2+2qq^2 + 2q and demand is 8q8q, set q2+2q=8qq^2 + 2q = 8q, so q2−6q=0q^2 - 6q = 0, giving q(q−6)=0q(q - 6) = 0. Solutions are q=0q = 0 or q=6q = 6; equilibrium is at q=6q = 6 units.

💡 Remember: "Set equal, factor, pick the positive quantity!"

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5. Applications: Finding equilibrium points where supply-demand quadratics meet

Applications: Finding Equilibrium Points Where Supply-Demand Quadratics Meet

In economics, equilibrium occurs where supply equals demand. When supply S(p)S(p) and demand D(p)D(p) are quadratic functions of price pp, setting S(p)=D(p)S(p) = D(p) yields a quadratic equation whose solutions represent equilibrium prices.

Factoring simplifies solving these equations and interpreting multiple equilibria.

Core Rules:

  • Set supply equal to demand: S(p)=D(p)S(p) = D(p)
  • Rearrange to standard form: ap2+bp+c=0ap^2 + bp + c = 0
  • Factor the quadratic (if possible) to find equilibrium prices directly
  • Each positive real root represents a feasible equilibrium price; negative or complex roots are economically meaningless

Multiple equilibria indicate market instability or multiple stable states.

Example: Suppose supply is S(p)=p2+2pS(p) = p^2 + 2p and demand is D(p)=15D(p) = 15. Setting p2+2p=15p^2 + 2p = 15 gives p2+2p−15=0p^2 + 2p - 15 = 0. Factoring: (p+5)(p−3)=0(p+5)(p-3)=0, so p=−5p=-5 or p=3p=3. Only p=3p=3 dollars is economically valid.

TASK_1[0 / 3]
LVL_2
EXEC: ALGORITHM

Suppose the supply function for a product is S(p)=p2+5pS(p) = p^2 + 5p and the demand function is D(p)=14D(p) = 14. Find the economically valid equilibrium price pp in dollars.

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