solution to Example 1 of Factoring Special Products :

Let P = 16x^{2}y^{2} + 24xyz + 9z^{2} = (4xy)^{2} + 2(4xy)(3z) + (3z)^{2}

This is like a^{2} + 2ab + b^{2} with (4xy) in place of a and 3z in place of b

We have a^{2} + 2ab + b^{2} = (a + b)^{2} [ See Formula 1]

∴ P = (4xy + 3z)^{2}

Thus Algebra Factoring of 16x^{2}y^{2} + 24xyz + 9z^{2} by using Algebra Formulas gave the factors as (4xy + 3z)^{2}. Ans.

Thus the problem on Factoring Expressions is solved.

Example 2 of Factoring Expressions :

Factorize (2a + 3b)^{2} - 2(2a + 3b)(a - b) + (a - b)^{2}

solution to Example 2 of Factoring Special Products :

Let P = (2a + 3b)^{2} - 2(2a + 3b)(a - b) + (a - b)^{2} = (first term)^{2} - 2(first term)(second term) + (second term)^{2} = (first term - second term)^{2} [See Formula 2]

= {(2a + 3b) - (a - b)}^{2} = {2a + 3b - a + b}^{2} = {a + 4b}^{2}

Thus Algebra Factoring of (2a + 3b)^{2} - 2(2a + 3b)(a - b) + (a - b)^{2} by using Algebra Formulas gave the factors as {a + 4b}^{2} Ans.

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solution to Example 3 of Factoring Special Products :

Let P = 16x^{4} + 4x^{2} + 1= (4x^{2})^{2} + 1^{2} + (4x^{2})(1) This looks like a^{2} + b^{2} + ab with (4x^{2}) in place of a and 1 in place of b To make it (a + b)^{2}, we have to add one more ab { = (4x^{2})(1)} [See Formula 1] Not to change the value, let us add and subtract. Adding and subtracting (4x^{2})(1) to the R.H.S. of P, we get P = (4x^{2})^{2} + 1^{2} + (4x^{2})(1) + (4x^{2})(1) - (4x^{2})(1) = (4x^{2})^{2} + 1^{2} + 2(4x^{2})(1) - (4x^{2})(1) = (4x^{2} + 1)^{2} - (2x)^{2} This is difference of squares of two terms which is equal to the product of the sum and difference of the two terms [See Formula 3]. ∴ P = {(4x^{2} + 1) + (2x)}{(4x^{2} + 1) - (2x)} = (4x^{2} + 2x + 1)(4x^{2} - 2x + 1).

Thus Algebra Factoring of 16x^{4} + 4x^{2} + 1 by using Algebra Formulas gave the factors as (4x^{2} + 2x + 1)(4x^{2} - 2x + 1). Ans.

solution to Example 4 of Factoring Special Products :

Let P = 27x^{3} + 27x^{2} + 9x + 1= (3x)^{3} + 3(3x)^{2}(1) + 3(3x)(1)^{2} + (1)^{3} This looks like a^{3} + 3a^{2}b + 3ab^{2} + b^{3} with (3x) in place of a and 1 in place of b. We have a^{3} + 3a^{2}b + 3ab^{2} + b^{3} = a^{3} + 3ab(a + b) + b^{3} = (a + b)^{3} [See Formula 6] Applying this Formula here, we get P = (3x + 1)^{3}.

Thus Algebra Factoring of 27x^{3} + 27x^{2} + 9x + 1 by using Algebra Formulas gave the factors as (3x + 1)^{3}. Ans.

Exercise : Factoring Expressions

Do the following problems on Factoring Expressions.

Factorize (ab + cd)^{2} + 18(ab + cd) + 81

Factorize 49x^{2} - 28xyz + 4y^{2}z^{2}

Factorize 625a^{4} + 25a^{2}b^{2} + b^{4}

Factorize 8a^{3} + 60a^{2} + 150a + 125

For Answers to these problems on Factoring Expressions See at the bottom of the Page.

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