Question

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

Answer 1

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

Step-by-step 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

$7=c(4^{3}+2(4)^{2}-5(4)-6)\\7=c(64+32-20-6)=70c\\c=\frac{1}{10}$

∴Required expression is $\frac{1}{10}(x^{3}+2x^{2}-5x-6)$