document.write( "Question 1038159: The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?
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Algebra.Com's Answer #652883 by Boreal(15235) $\"\"$ $\"About$

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ax^3+bx^2+cx+d is the polynomial
\n" ); document.write( "-a+b-c+d=15
\n" ); document.write( "d=0, so one intercept is (0,0). The reason d=0 is that f(0)=0. When f(0) is applied to the function, all the x terms disappear, since they are all zero. The only term left is d, the constant. But d is 0, because f(0)=0.
\n" ); document.write( "a+b+c=-5
\n" ); document.write( "8a+4b+2c=12
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\n" ); document.write( "-a+b+c=15. This comes from the first equation, removing d, since it is 0. I can then add that and the one below it.
\n" ); document.write( "a+b+c=-5
\n" ); document.write( "2b+2c=10
\n" ); document.write( "b+c=5, dividing by 2.
\n" ); document.write( "a=-10
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\n" ); document.write( "-80+10+2b=12, because 2b+2c+10
\n" ); document.write( "2b=82
\n" ); document.write( "b=41
\n" ); document.write( "c=-36
\n" ); document.write( "-10x^3+41x^2-36x=0
\n" ); document.write( " $\"graph%28300%2C200%2C-10%2C10%2C-20%2C20%2C-10x%5E3%2B41x%5E2-36x%29\"$
\n" ); document.write( "factor out an -x
\n" ); document.write( "-x(10x^2-41x+36)=0
\n" ); document.write( "x=0
\n" ); document.write( "x=(1/20)(41 +/-sqrt (1681-1440); sqrt 241=15.52
\n" ); document.write( "x=56.52/20, or 2.83
\n" ); document.write( "x=25.48/20, or 1.274
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