Question

Numer 9 and 10. Need Solutions​

Answer 1

QUESTION 9

The given function has x-intercepts at;

$x=-2$ with multiplicity 1.

$x=0$ with multiplicity even, say 2.

$x=3$ with multiplicity 1.

By the factor theorem; $x+2,(x-0)^2,x-3$ are factors of the polynomial function.

The possible formula for the graph is

$p(x)=ax^2(x+2)(x-3)$

The point (-1,4) lies on this graph

$4=a(-1)^2(-1+2)(-1-3)$

$4=-4a$

$a=-1$

Hence a possible formula is

$p(x)=-x^2(x+2)(x-3)$

QUESTION 10

The given polynomial function has x-intercept at x=-2, with and odd multiplicity, say 1.

It was given that;

$p(i)=0$

This implies that $x=i$ is a solution.

By the complex conjugate property, $x=-i$ is also a solution.

By the factor theorem;

$P(x)=a(x+2)(x-i)(x+i)$

Apply difference of two squares and simplify to get;

$P(x)=a(x+2)(x^2+1)$

The graph passes  through (2,-4).

$-4=a(2+2)(2^2+1)$

$-4=20a$

$a=-\frac{1}{5}$

A possible formula is

$P(x)=-\frac{1}{5}(x+2)(x^2+1)$