document.write( "Question 24971: Factor completely:
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\n" ); document.write( "1)6(2p+q)²-5(2P+q)-25=\r
\n" ); document.write( "\n" ); document.write( "Factor. Assume that variables in exponents represent natural numbers.
\n" ); document.write( "1) 25y^2m - (x²ⁿ-2xⁿ+1)\r
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Algebra.Com's Answer #13354 by AnlytcPhil(1806) $\"\"$ $\"About$
You can put this solution on YOUR website!
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document.write( "y³ - 3y² - 4y + 12 \r\n" );
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document.write( "Factor the first two terms only:\r\n" );
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document.write( "y²(y - 3) - 4y + 12\r\n" );
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document.write( "Factor the last two terms only, taking out -4, remembering\r\n" );
document.write( "that when you take a negative like -4 out of a positive like +12,\r\n" );
document.write( "you get a negative (like -3).\r\n" );
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document.write( "y²(y - 3) - 4(y - 3)\r\n" );
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document.write( "I will color the like factors red\r\n" );
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document.write( "y²(y - 3) - 4(y - 3)\r\n" );
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document.write( "Thatke out the red factor (y - 3)\r\n" );
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document.write( "(y - 3)(y² - 4)\r\n" );
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document.write( "Now the second parenthetical expression will factor as the\r\n" );
document.write( "difference of two squares\r\n" );
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document.write( "(y - 3)(y² - 2²)\r\n" );
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document.write( "(y - 3)(y - 2)(y + 2) \r\n" );
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document.write( "2x³ + 6x² - 8x - 24\r\n" );
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document.write( "First factor 2 out of the entire expression:\r\n" );
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document.write( "2[x³ + 3x² - 4x - 12]\r\n" );
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document.write( "Then factor the bracket as in the previous problem:\r\n" );
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document.write( "Factor x² out of first two terms in brackets:\r\n" );
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document.write( "2[x²(x + 3) - 4x - 12]\r\n" );
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document.write( "Factor -4 out of last two terms in brackets, remembering that\r\n" );
document.write( "when you take a negative like -4 out of another negative, like -12,\r\n" );
document.write( "you get a positive +3 :\r\n" );
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document.write( "2[x²(x + 3) - 4(x + 3)]\r\n" );
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document.write( "I'll color the like factors red:\r\n" );
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document.write( "2[x²(x + 3) - 4(x + 3)]\r\n" );
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document.write( "Take out the red factor from the terms inside the bracket:\r\n" );
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document.write( "2[(x + 3)(x² - 4)]\r\n" );
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document.write( "Now the second binomial will factor as the difference \r\n" );
document.write( "of two squares\r\n" );
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document.write( "2[(x + 3)(x² - 2²)]\r\n" );
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document.write( "2[(x + 3)(x - 2)(x + 2)]\r\n" );
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document.write( "Dispense with the brackets\r\n" );
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document.write( "2(x + 3)(x - 2)(x + 2) \r\n" );
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document.write( "6(2p+q)² - 5(2p+q) - 25 \r\n" );
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document.write( "6(2p+q)² - 5(2p+q) - 25\r\n" );
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document.write( "You can now treat the red parenthetical expression just as you\r\n" );
document.write( "would treat a single letter.  That is, you can factor the above\r\n" );
document.write( "just as you would factor 4x² - 5x - 25 as (3x + 5)(2x - 5), but\r\n" );
document.write( "use brackets:\r\n" );
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document.write( "6(2p+q)² - 5(2p+q) - 25\r\n" );
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document.write( "[3(2p+q) + 5][2(2p+q) - 5]\r\n" );
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document.write( "Now remove the parentheses inside the brackets:\r\n" );
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document.write( "[6p + 3q + 5][4p + 2q - 5]\r\n" );
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document.write( "Change the brackets to parentheses:\r\n" );
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document.write( "(6p + 3q + 5)(4p + 2q - 5)\r\n" );
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document.write( "===============================================\r\n" );
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document.write( "25y^2m - (x²ⁿ-2xⁿ+1)\r\n" );
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document.write( "Write the first term as (5y^m)².\r\n" );
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document.write( "(5y^m)² - (x²ⁿ-2xⁿ+1)\r\n" );
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document.write( "Write the second expression as [(xⁿ)2-2(xⁿ)+1]\r\n" );
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document.write( "(5y^m)² - [(xⁿ)²-2(xⁿ)+1]\r\n" );
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document.write( "(5y^m)² - [(xⁿ)2-2(xⁿ)+1]\r\n" );
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document.write( "Now factor the bracketed expression treating the red parentheses\r\n" );
document.write( "as though they were just a single letter, but we'll need to go\r\n" );
document.write( "to braces\r\n" );
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document.write( "(5y^m)² - {(xⁿ)2-2(xⁿ)+1}\r\n" );
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document.write( "(5y^m)² - {[(xⁿ)-1][(xⁿ)-1]}\r\n" );
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document.write( "Since the two bracketed factors in the braces are the same we \r\n" );
document.write( "can just write\r\n" );
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document.write( "(5y^m)² - (xⁿ-1)2\r\n" );
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document.write( "(5y^m)² - (xⁿ-1)²\r\n" );
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document.write( "Now to do some more coloring:\r\n" );
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document.write( "(5y^m)² - (xⁿ-1)²\r\n" );
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document.write( "Now this is the difference of two squares and factors as\r\n" );
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document.write( "[(5y^m) - (xⁿ-1)][(5y^m) + (xⁿ-1)]\r\n" );
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document.write( "Remove the parentheses inside the bracket:\r\n" );
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document.write( "[5y^m - xⁿ + 1][5y^m + xⁿ - 1]\r\n" );
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document.write( "Change the brackets to parentheses:\r\n" );
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document.write( "(5y^m - xⁿ + 1)(5y^m + xⁿ - 1)\r\n" );
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document.write( "I'm not going to color on the rest.  See if you can figure \r\n" );
document.write( "them out without colors:\r\n" );
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document.write( "x^6a  - t^3b\r\n" );
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document.write( "Write these terms as\r\n" );
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document.write( "(x^2a)³ - (t^b)³\r\n" );
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document.write( "This is the difference of two cubes.  Use the rule for factoring\r\n" );
document.write( "the sum or difference of two cubes:\r\n" );
document.write( "                                                _\r\n" );
document.write( "                           P³ + Q³ = (P + Q)(P² + PQ + Q²)\r\n" );
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document.write( "[(x^2a) - (t^b)][(x^2a)² + (x^2a)(t^b) + (t^b)²] \r\n" );
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document.write( "[x^2a - t^b][x^4a + x^2at^b + t^2b]\r\n" );
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document.write( "(y-1)⁴ - (y-1) \r\n" );
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document.write( "Factor out (y-1)\r\n" );
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document.write( "(y-1)[(y-1)³ - 1]\r\n" );
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document.write( "(y-1)[(y-1)³ - 1³]\r\n" );
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document.write( "The bracketed expression is the difference of two cubes, and so we\r\n" );
document.write( "use the rule I gave in the last problem:\r\n" );
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document.write( "(y-1)[(y-1) - 1][(y-1)² + (y-1)(1) + 1²]\r\n" );
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document.write( "Remove all the parentheses inside the brackets:\r\n" );
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document.write( "(y-1)[y - 1 - 1][y²-2y+1 + y-1 + 1]\r\n" );
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document.write( "(y-1)[y-2}[y²-y+1]\r\n" );
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document.write( "(y-1)(y-2)(y²-y+1)\r\n" );
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document.write( "=============================================\r\n" );
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document.write( "X⁶ - 2X⁵ + X⁴ - X² + 2X - 1\r\n" );
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document.write( "Factor X⁴ out of the first three terms:\r\n" );
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document.write( "X⁴(X²-2X+1) - X² + 2X - 1\r\n" );
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document.write( "Factor -1 out of the last three terms, remembering to change\r\n" );
document.write( "the sign when factoring out a negative like -1\r\n" );
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document.write( "X⁴(X²-2X+1) - 1(X²-2X+1)\r\n" );
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document.write( "Now factor out the common factor (X²-2X+1)\r\n" );
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document.write( "(X²-2X+1)[X⁴ - 1]\r\n" );
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document.write( "The first factor factors as (X-1)(X-1) = (X-1)²\r\n" );
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document.write( "(X-1)²[X⁴ - 1]\r\n" );
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document.write( "The second factor is the difference of two squares:\r\n" );
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document.write( "(X-1)²[(X²)² - 1²]\r\n" );
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document.write( "(X-1)²[(X²-1)(X²+1)]\r\n" );
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document.write( "The first factor in the brackets is the difference of two squares\r\n" );
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document.write( "(X-1)²[(X-1)(X+1)(X²+1)]\r\n" );
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document.write( "Dispense with the brackets\r\n" );
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document.write( "(X-1)²(X-1)(X+1)(X²+1)\r\n" );
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document.write( "The first two factors have the same base (X-1), so we can write them\r\n" );
document.write( "as the cube of this:\r\n" );
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document.write( "(X-1)³(X+1)(X²+1)\r\n" );
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document.write( "Edwin\r\n" );
document.write( "AnlytcPhil@aol.com
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