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How do you factor trinomials and binomials on STAAR, including the GCF, the difference of two squares, and perfect-square trinomials?

Factor, if possible, trinomials with real factors in the form ax2+bx+cax^2 + bx + c, including perfect-square trinomials, and decide if a binomial is a difference of two squares and rewrite it (TEKS A.10E, A.10F).

A STAAR Algebra I answer on factoring trinomials of the form ax squared plus bx plus c, perfect-square trinomials, and the difference of two squares (TEKS A.10E, A.10F), the GCF-first routine, and the reference-sheet identities.

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  1. What this topic is asking
  2. Step one is always the GCF
  3. Difference of two squares
  4. Perfect-square trinomials
  5. Trinomials with a leading coefficient
  6. How STAAR examines factoring
  7. Reading the signs of a trinomial
  8. Try this

What this topic is asking

Factoring is the bridge from polynomial arithmetic to solving quadratics, and TEKS A.10E and A.10F sit in the Number and Algebraic Methods category. STAAR expects you to factor trinomials of the form ax2+bx+cax^2 + bx + c with real factors (including perfect squares) and to recognize and rewrite a difference of two squares. The same skill reappears in the Quadratic Functions and Equations category when you solve ax2+bx+c=0ax^2 + bx + c = 0 by factoring, so this is high-value work.

Step one is always the GCF

Before any pattern, pull out the greatest common factor of all terms. It is the largest number and variable power dividing every term.

6x39x2+3x=3x(2x23x+1).6x^3 - 9x^2 + 3x = 3x(2x^2 - 3x + 1).

Skipping the GCF is the top reason a "complete factorization" answer is wrong: 4x2364x^2 - 36 must become 4(x29)4(x^2 - 9) before the difference of squares is applied. On a multiple-choice item, a partially factored choice like 4(x29)4(x^2 - 9) is a deliberate distractor.

Difference of two squares

A binomial that is one square minus another square factors by the reference-sheet identity:

A sum of squares such as x2+9x^2 + 9 does not factor over the real numbers, so do not force it. Recognize the squares: 4949, 36x236x^2, 14\frac{1}{4}, and so on.

Perfect-square trinomials

When a trinomial is the square of a binomial, the reference sheet gives the pattern:

a2+2ab+b2=(a+b)2,a22ab+b2=(ab)2.a^2 + 2ab + b^2 = (a + b)^2, \qquad a^2 - 2ab + b^2 = (a - b)^2.

The test is whether the first and last terms are perfect squares and the middle term is twice their roots. For x2+10x+25x^2 + 10x + 25: x2=x\sqrt{x^2} = x, 25=5\sqrt{25} = 5, and 2x5=10x2 \cdot x \cdot 5 = 10x matches, so it is (x+5)2(x + 5)^2.

Trinomials with a leading coefficient

For x2+bx+cx^2 + bx + c, find two numbers with product cc and sum bb. For ax2+bx+cax^2 + bx + c with a1a \neq 1, use the AC method.

How STAAR examines factoring

  • Multiple choice and multiselect. Pick the complete factorization, or select all expressions equivalent to a given polynomial. The GCF and the "stop too early" distractors are standard.
  • Drag and drop. Drag the correct constants into binomial templates such as (x+)(x+)(x + \square)(x + \square), which tests the product-and-sum reasoning directly.
  • Inside quadratic items. Factoring is the fastest route to the zeros of a parabola and the solutions of ax2+bx+c=0ax^2 + bx + c = 0, so it threads through Reporting Category 4.

A clarifying check is that factoring undoes multiplying: expanding your factors must rebuild the original polynomial. If it does not, a sign or a number is off.

Reading the signs of a trinomial

The signs in x2+bx+cx^2 + bx + c tell you the signs of the factors before you search. If cc is positive, the two numbers share a sign (both positive when bb is positive, both negative when bb is negative), as in x27x+12=(x3)(x4)x^2 - 7x + 12 = (x - 3)(x - 4). If cc is negative, the two numbers have opposite signs, and the larger absolute value carries the sign of bb, as in x2x12=(x4)(x+3)x^2 - x - 12 = (x - 4)(x + 3). Using the sign rule first narrows the candidate pairs immediately and prevents the common error of producing factors whose product has the wrong constant sign. Combined with the GCF-first habit and the two reference-sheet identities, this gives a complete, ordered procedure that handles every factorable polynomial the test puts in front of you.

Try this

Q1. Factor completely: 3x2273x^2 - 27. [1 point]

  • Cue. GCF first: 3(x29)=3(x3)(x+3)3(x^2 - 9) = 3(x - 3)(x + 3).

Q2. Factor: x212x+36x^2 - 12x + 36. [1 point]

  • Cue. Perfect square: 36=6\sqrt{36} = 6 and 26=122 \cdot 6 = 12, so (x6)2(x - 6)^2.

Exam-style practice questions

Practice questions written in the style of TEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

STAAR (style)1 marksMultiple choice. Which expression is the complete factorization of 4x2364x^2 - 36? (A) 4(x29)4(x^2 - 9) (B) (2x6)(2x+6)(2x - 6)(2x + 6) (C) 4(x3)(x+3)4(x - 3)(x + 3) (D) 4(x9)(x+9)4(x - 9)(x + 9)
Show worked answer →

The correct answer is (C).

Always take the greatest common factor first: 4x236=4(x29)4x^2 - 36 = 4(x^2 - 9). The factor x29x^2 - 9 is a difference of two squares (the reference sheet gives a2b2=(ab)(a+b)a^2 - b^2 = (a - b)(a + b)), so x29=(x3)(x+3)x^2 - 9 = (x - 3)(x + 3). The complete factorization is 4(x3)(x+3)4(x - 3)(x + 3). Choice (A) stops too early and choice (B) leaves a common factor of 2 inside each binomial, so neither is fully factored.

STAAR (style)2 marksDrag and drop. A student factors x2+7x+12x^2 + 7x + 12. Drag the two binomial factors into the boxes: (x+)(x+)(x + \square)(x + \square). Available tiles: 2,3,4,62, 3, 4, 6.
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The factors are (x+3)(x+4)(x + 3)(x + 4), so the tiles 3 and 4 go in the boxes.

For x2+bx+cx^2 + bx + c with leading coefficient 1, find two numbers that multiply to c=12c = 12 and add to b=7b = 7. The pairs multiplying to 12 are 1×121 \times 12, 2×62 \times 6, and 3×43 \times 4; only 3+4=73 + 4 = 7. So x2+7x+12=(x+3)(x+4)x^2 + 7x + 12 = (x + 3)(x + 4). The tile pair 22 and 66 multiplies to 12 but adds to 8, the distractor; check the sum, not just the product.

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