Question

Third-degree, with zeros of -3,-1 and 2, and passes through the point (4,7)

Answer 1

Required Expression is
`\frac{\textbf{1}}{\textbf{10}}(\textbf{x}^{\textbf{3}}\textbf{+}\textbf{2x}^{\textbf{2}}\textbf{-}\textbf{5x}\textbf{-}\textbf{6})`

Explanation:

Assuming that the question is to find a third degree expression
`f(x)` that has zeros
`-3,-1,2` and the equation
`y=f(x)` passes through
`(4,7)`,

If the roots/zeroes of a
`n^(th)` order expression are given as
`r_(1),r_(2),r_(3)..... r_(n)`, the expression is given by
`f(x)=c(x-r_(1))(x-r_(2))(x-r_(3)).....(x-r_(n))`.

Since we know the three roots of the third degree expression, the function is

`f(x)=c(x-(-3))(x-(-1))(x-2)=c(x+3)(x+1)(x-2)=c(x^(3)+2x^(2)-5x-6)`

Also,
`y=f(x)` passes through
`(4,7)`, so

`7=c(4^(3)+2(4)^(2)-5(4)-6)\\7=c(64+32-20-6)=70c\\c=(1)/(10)`

∴Required expression is
`(1)/(10)(x^(3)+2x^(2)-5x-6)`