document.write( "Question 1018196: NEED ANSWERS ASAP
\n" ); document.write( "2. Near the beginning of Lesson 5.3, a strategy for factoring trinomials of the form x2 + bx + c was developed by exploring the product of the binomials (x + p) and (x + q).
\n" ); document.write( "/1 a. Explain how the development of this factoring strategy is an example of working backwards to solve a problem.
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\n" ); document.write( " b. The product of (x + p)(x + q) can be written as x2 + (p + q)x + pq.
\n" ); document.write( "/1 i. An intermediate step in this multiplication is
\n" ); document.write( " x2 + px + qx + pq = x2 + (p + q)x + pq.
\n" ); document.write( " Explain why px + qx = (p + q)x.
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\n" ); document.write( "\n" ); document.write( "/1 ii. Explain why the expression x2 + (p + q)x + pq leads to the need to determine integers that add to b and have a product c when factoring a trinomial of the form x2 + bx + c.\r
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Algebra.Com's Answer #634362 by josgarithmetic(39618) $\"\"$ $\"About$

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Perform the multiplication steps and use Distributive Property. These steps will speak for themselves. The simplification gives a formula that can be used. The simplification part is the combining of terms of x. Looking at the steps BEFORE the combining of the like terms of x, you see the sum of numbers which you would be looking for in an attempt to factorize the trinomial expression. \n" ); document.write( "

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