Question

find the zeros of the polynomial p[x]=3x-18

Answer 1

Answer:

p(x)=x

3

−2x

2

+3x−18 is not a multiple. of x-3x−3 .

Step-by-step explanation:

If the question asks us to show that p(x)={x}^3-{2x}^2+3x-18p(x)=x

3

−2x

2

+3x−18 is the multiple of x-3x−3 , that in another words is proving that x-3x−3 is a factor of the the polynomial p(x)=....p(x)=.... . So let's use Factor Theorem to solve.

Let's find the zero of x-3x−3

x-3=0x−3=0

x=0+3x=0+3

x=3x=3

3 is the zero of the polynomial.

p(x)={x}^3-{2x}^2+3x-18p(x)=x

3

−2x

2

+3x−18

p(3)={(3)}^3-{(2*3)}^2+3(3)-18p(3)=(3)

3

−(2∗3)

2

+3(3)−18

p(3)=27-36+9-18p(3)=27−36+9−18

p(3)= -18p(3)=−18

Answer 2

Finding a zero of p(x), is same as solving the equation.

• p(x) = 0

$\large\bf{\green{ \bf \large \longrightarrow \: }} \rm \large \: 3x \: - \: 18 \: = \: 0$

$\large\bf{\green{ \bf \large \longrightarrow \: }} \rm \large \:3x \: = \: 18$

$\large\bf{\green{ \bf \large \longrightarrow \: }} \rm \large \:x \: = \: \frac{18}{6}$

$\large\bf{\green{ \bf \large \longrightarrow \: }} \rm \large \:x \: = \: \cancel\frac{{18} }{6} \: \: ^{3}$

$\large\bf{\green{ \bf \large \longrightarrow \: }} \rm \large \:x \: = \: 3$

⇛ So , 3 is a zero of the polynomial

3x - 18