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Dynamic tasks on square identities, difference of squares, common factors and polynomial factorisation.
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Factor out x. A product is zero when at least one factor is zero. Find both solutions and add them.
Square of a sum: (x + a)² = x² + 2ax + a². The middle term comes from adding the two cross products ax. Thus the x coefficient is 2a; x² + a² alone is incomplete.
Divide by the coefficient of x². Find both solutions with opposite signs. The distance is the larger solution minus the smaller one.
Substitute the given solution for x. Square it first, multiply by the coefficient, then isolate c in the resulting equality.
A difference of squares factors as a² − b² = (a − b)(a + b). Apply this to the given bases. Expanding the product cancels the cross terms.
Square of a sum: (x + a)² = x² + 2ax + a². The middle term comes from adding the two cross products ax. Thus the x coefficient is 2a; x² + a² alone is incomplete.
Find two numbers whose sum is s and product is p. If they are a and b, then x² + sx + p = (x + a)(x + b). Check both conditions.
Find a factor common to all terms. Divide each term by it and write the quotients in parentheses: ax + ab = a(x + b). Check by expanding.
Add the removed area to the remaining area. Solve x² = S. Only the positive solution can represent a side length.
Square of a difference: (x − a)² = x² − 2ax + a². The middle term is negative and the final term positive because (−a) × (−a) = a².
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The identities (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a − b)(a + b) help expand and factor expressions efficiently. The middle term of a squared binomial is always twice the product.
First take out any common factor. Then look for a difference of squares or a perfect-square trinomial. Check the result by multiplying the factors back out; sign errors usually appear in the middle term.
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https://www.catestia.com/en/mathematics/8th-grade/algebraic-identities-and-factorisation-test/65/