document.write( "Question 1135164: Write an equation for the third degree polynomial, (-1,0),(2,0) also f(0)=-34 \n" ); document.write( "

Algebra.Com's Answer #752733 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "The solution by tutor @mathsolverplus is incomplete; the general form of a cubic equation has a linear term. His polynomial is one that passes through the three given points and has no linear term.

\n" ); document.write( "There are an infinite number of third degree polynomials that pass through the three given points (-1,0), (2,0) and (0,-34).

\n" ); document.write( "The general form of a third degree polynomial in factored form is

\n" ); document.write( " $\"y+=+a%28x-p%29%28x-q%29%28x-r%29\"$

\n" ); document.write( "where p, q, and r are the three roots.

\n" ); document.write( "We know that two of the roots are -1 and 2, so the general form of this third degree polynomial is

\n" ); document.write( " $\"y+=+a%28x%2B1%29%28x-2%29%28x-r%29\"$

\n" ); document.write( "The other piece of information we have is that f(0)=-34. Using that in our general form gives us

\n" ); document.write( " $\"-34+=+a%281%29%28-2%29%28-r%29\"$
\n" ); document.write( " $\"-34+=+2ar\"$
\n" ); document.write( " $\"r+=+-34%2F%282a%29\"$

\n" ); document.write( "This shows us that the third root depends on the value of the leading coefficient; and that means there are an infinite number of polynomials whose graphs pass through the three given points.

\n" ); document.write( "The graph below shows plots of two of those functions:
\n" ); document.write( "(1) a=1 --> r = -17; y = (x+1)(x-2)(x+17) (red graph)
\n" ); document.write( "(2) a=3 --> r = -17/3; y = 3(x+1)(x-2)(x+17/3) (green graph)

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