Question

Show that -3/2 and -3 are zeros of the polynomial 6x³+23x²+9x-18 also find the third zero of the polynomial

Answer 1

check out pic for ur answer

Answer 2

Let p(x) be the given polynomial 6x³ +23x² + 9x - 18, then p(x) = 6x³ +23x² + 9x - 18

$\sf{Now,p}$$\sf\frac{(-3)}{(2)}^3+23$$\sf\frac{(-3)}{(2)}^2+9$$\sf\frac{(-3}{(2)}-18$

= $\sf{6}$$\sf\frac{(-27)}{(8)}+23$$\sf\frac{(9)}{(4)}-\frac{27}{2}-18$

= $\sf\frac{-81}{4}+\frac{207}{4}-\frac{54}{4}-18$

= $\sf\frac{-81+207-54}{4}-18$

= $\sf\frac{207-135}{4}-18$

= $\sf\frac{72}{4}-18$

= $\sf{18-18}$

= $\sf{0}$

$\implies$$\sf\frac{-3}{2}$

is a zero of the given polynomial 6x³ + 23x² + 9x - 18

and $\sf{p(-3)=6(-3)^3+23(-3)^3+9(-3)-18}$

= $\sf{6(-27)+23(9)-27-18}$

= $\sf{-162+207-45}$

= $\sf{-207-207}$

= $\sf{0}$

$\implies$ -3 is a zero of the given polynomial 6x³ + 23x² + 9x - 18

Since $\sf\frac{-3}{2}$ and $\sf{-3}$ are the zeros of the polynomial 6x³ + 23x² + 9x - 18, therefore both $\sf{(x)+}$$\sf\frac{(3)}{(2)}$ and (x+3) are the factors of 6x³ + 23x² + 9x - 18.

$\sf{Let\:g(x)=}$$\sf{(x)+}$$\sf\frac{(3)}{(2)}$$\sf{(x+3)}$=$\sf{(2x+3)(x+3)=2x^2+9x+9}$

• Since $\sf{(x)+}$$\sf\frac{(3)}{(2)}$ and (x+3) are the factors of p(x), therefore g(x) = 2x² + 9x + 9 is a factor of p(x).

So, when we divide p(x) by g(x) , we have

3x-2

Thus, 6x³ + 23x² + 9x - 18 = (2x² + 9x - 9)(3x - 2)

$\implies$The third factor of 6x³ + 23x² + 9x - 18 is 3x-2.