Recognise equivalent polynomial forms

At x = 0, (x + 2)² and x² + 4 both equal 4. Does this prove they are equivalent?

Move x from −2 to 2. Compare the actual square with the expression that omits the cross-products

What does their agreement at x = 0 show?

Verify the whole polynomial

Read the coefficient of each power and the constant term. Reordering terms does not change an expression; changing a coefficient changes the polynomial

1. 1Same expression, different form Equivalent polynomial forms have the same value for every real input. Expanding every product proves the identity
2. 2Keep every coefficient Coefficients include their signs. A missing power has coefficient zero
3. 3A check is not a proof One disagreement disproves equivalence. Agreement at one input alone does not prove it
4. 4Name the requested form A factorised product may be equivalent, but an expansion must have its brackets multiplied out

Construction and verification are different tasks

The preceding four topics ask for coefficients in supplied expansion or factor templates. This bank first offers complete expressions to choose from, then asks for entries belonging to the chosen expression

1. 1Write complete forms too Practise the corresponding full expansions and factorisations on paper

A binomial times a trinomial

Form all six products before collecting powers. A cubic product follows the same distributive law as the earlier two-bracket products

1. 1First multiplier Multiply 2x by all three terms \(\displaystyle 2x^3+2x^2+6x\)
2. 2Second multiplier Multiply −1 by all three terms \(\displaystyle -x^2-x-3\)
3. 3Collect Add coefficients only when their powers match \(\displaystyle 2x^3+x^2+5x-3\)

Find the signed coefficient of x in the complete expansion

Two cubes, not the cube of a sum

Identify each cubed quantity, then use the matching identity. These expressions differ from (a+b)³ and (a−b)³, which generally include mixed terms

1. 1Sum of cubes The first factor uses a sum. The quadratic has the opposite middle sign \(\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)\)
2. 2Difference of cubes The first factor uses a difference. The quadratic has a positive middle product \(\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)\)
3. 3Verify the cancellation Multiply both factors. The squared-linear and linear-squared cross terms cancel in pairs; only the two cubes remain

Choose an equivalent factorisation of this difference of cubes

Complete the signed coefficient in the quadratic factor

Which expanded form results when the cross terms cancel?

A repeated quadratic is one whole factor

Let U=x²−x+2. Squaring U means multiplying all three terms by all three terms. It does not mean squaring the three terms separately

1. 1Square the whole expression Each term multiplies every term. Include each pair of different terms twice \(\displaystyle (a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc\)
2. 2Collect equal powers The squared linear term and the constant times quadratic term both contribute to the coefficient of x²
3. 3Reverse by checking every coefficient A plausible repeated factor is not enough. Its complete square must recover the original polynomial
4. 4This expansion Collect all six types of products, including the repeated cross-products \(\displaystyle x^4-2x^3+5x^2-4x+4\)

Find the signed coefficient of x³ after squaring the whole quadratic

Which repeated factor reproduces every coefficient?

Group an already visible repeated bracket

Treat a repeated bracket as one quantity. Combine its other multipliers without squaring the shared bracket or dropping a sign

1. 1Take out the common expression The common factor x−2 occurs once in each term \(\displaystyle (x-2)((3x+1)+(x-5))\)
2. 2Collect inside Add signed coefficients in the second bracket \(\displaystyle (x-2)(4x-4)\)
3. 3If complete integer factors are requested The second bracket still has a numerical common factor \(\displaystyle 4(x-2)(x-1)\)

Find the constant in the combined second bracket

Read the exact factor table being requested

Different products can be equivalent after factors are swapped or rescaled. In this bank, number the factors in the correct supplied option from left to right and retain its signs. A numerical outside factor counts as a factor; a squared repeated factor is listed once

1. 1Outside factor In 2(x²+x+3), factor 1 is the constant 2 and factor 2 has coefficients 3, 1 and 1 in increasing-power order \(\displaystyle 2(x^2+x+3)\)
2. 2Missing powers are zero The second factor x²+3 has no x term \(\displaystyle (x-2)(x^2+3)\)
3. 3Do not claim a unique algebraic appearance The table convention identifies the displayed product, not the only mathematically equivalent product

Read the coefficient of x in factor 2 of this displayed product

Substitute exactly after establishing equivalence

Substitution can check a result, but one agreement cannot prove a polynomial identity. Keep fractions exact until the final value. Compare the two forms only after correctly identifying them

1. 1Example at x=−1/2 The cubed value is −1/8, so the expression is −1+27 \(\displaystyle 8(-\frac12)^3+27=26\)
2. 2Same value from the factors The linear factor is 2 and the quadratic factor is 13 \(\displaystyle (2x+3)(4x^2-6x+9)=2(13)=26\)

Evaluate the expression exactly at the stated input

Equivalent factor form versus complete factorisation

The requested task matters. A valid product can still contain a factor that splits further. This mixed bank asks for an equivalent supplied factor form and its ordered coefficients. Such a product need not be completely factorised

1. 1Equivalent intermediate form The bracket x²−1 can still be factorised over the integers \(\displaystyle 2(x^2-1)=2(x-1)(x+1)\)
2. 2State the domain when necessary What counts as irreducible depends on the permitted coefficients. Here the core construction templates use the stated integer-coefficient conventions