document.write( "Question 722966: My daughter has a question on a chapter review that I cannot figure out. She is to factor and check 16x^2 + 81. If it were 16x^2-81, I could understand it as (4x+9)(4x-9) and that checks out, but I cannot seem to figure out the equation as it is written. Can you help me please? \n" ); document.write( "

Algebra.Com's Answer #442964 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
Using only real number, you cannot factor $\"16x%5E2+%2B+81\"$ ,
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\n" ); document.write( "As you said
\n" ); document.write( " $\"16x%5E2-81=%284x%2B9%29%284x-9%29\"$ can be factored, and when one or the other of those factors is zero (for x=9/4 or x=-9/4), their product is zero too, meaning $\"16x%5E2-81=0\"$ .
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\n" ); document.write( " $\"16x%5E2-72x%2B81=%284x-9%29%5E2\"$ or $\"16x%5E2%2B72x%2B81=%284x%2B9%29%5E2\"$ can be factored too, and become zero when they have a zero factor.
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\n" ); document.write( "If you could factor $\"16x%5E2+%2B+81\"$ using real numbers, you would be able to make it zero for some real value of x, and $\"16x%5E2+%2B+81=0\"$ would have a solution.
\n" ); document.write( "However, for real values of x, $\"16x%5E2%3E=0\"$ and $\"16x%5E2+%2B+81%3E=81\"$ cannot be zero.
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\n" ); document.write( "Maybe your daughter is studying complex numbers, or maybe she is supposed to say it cannot be factored, or maybe there was a typo.
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\n" ); document.write( "You can factor $\"16x%5E2+%2B+81\"$ using complex numbers, including the imaginary number $\"i\"$ , whose square is $\"-1\"$
\n" ); document.write( " $\"%284x-9i%29%284x%2B9i%29=16x%5E2-81%2Ai%5E2=16x%5E2-81%28-1%29=16x%5E2%2B81\"$ \n" ); document.write( "

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