Question

$f(x)=2x^3+6x^2+4x$
$g(x)=x^2+3x+2$
The polynomials f(x) and g(x) are defined above. Which of the following polynomials is divisible by
$2x+3?$
$A) h(x) = f(x) + g(x)$
$B) p(x) = f(x) + 3g(x)$
$C) r(x) = 2f(x) + 3g(x)$
$D) s(x) = 3f(x) + 2g(x)$​

Answer 1

ANSWER:

Given:

• f(x) = 2x³ + 6x² + 4x
• g(x) = x² + 3x + 2

To Find:

• Which polynomial is divisible be 2x + 3

Concept:

• First we will find the zero of 2x + 3, and then put the value in f(x) and g(x) one by one. And then we will put the values of f(x)and g(x) in the given options. And the option which will yield 0 as the answer, will be our answer.

Solution:

Firstly we find zero of 2x + 3.

So,

⇒ 2x + 3 = 0

⇒ 2x = -3

⇒ x = -3/2----(1)

Now we put value of x from (1) in f(x) :

$\sf{f(x) = 2x^3 + 6x^2 + 4x} \\\\:\implies \sf{f\left(\dfrac{-3}{2}\right) = 2*\left(\dfrac{-3}{2}\right)^3 + 6*\left(\dfrac{-3}{2}\right)^2 + 4*\left(\dfrac{-3}{2}\right)} \\\\:\implies \sf{f\left(\dfrac{-3}{2}\right) = 2*\left(\dfrac{-27}{8}\right) + 6*\left(\dfrac{9}{4}\right) + 4*\left(\dfrac{-3}{2}\right)} \\\\:\implies \sf{f\left(\dfrac{-3}{2}\right) = \dfrac{-27}{4} + \dfrac{27}{2} - 6 }\\\\:\implies \sf{f\left(\dfrac{-3}{2}\right) = \dfrac{-27}{4}+ \dfrac{54}{4} - \dfrac{24}{4}} \\\\:\implies \sf{f\left(\dfrac{-3}{2}\right) = \dfrac{3}{4}----(2)}$

Now we put value of x from (1) in g(x) :

$\sf{g(x) = x^2 + 3x + 2} \\\\:\implies \sf{g\left(\dfrac{-3}{2}\right) = \left(\dfrac{-3}{2}\right)^2 + 3*\left(\dfrac{-3}{2}\right) + 2} \\\\:\implies \sf{g\left(\dfrac{-3}{2}\right) = \dfrac{9}{4} + 3*\left(\dfrac{-3}{2}\right) + 2} \\\\:\implies \sf{g\left(\dfrac{-3}{2}\right) = \dfrac{9}{4} - \dfrac{9}{2} + 2} \\\\:\implies \sf{g\left(\dfrac{-3}{2}\right) = \dfrac{9}{4} - \dfrac{18}{4} + \dfrac{8}{4}} \\\\:\implies \sf{g\left(\dfrac{-3}{2}\right) = \dfrac{-1}{4} -----(3)}$

Finally, we will put values of f(x) and g(x) in the following equations from (2) & (3) to get our answer:

$\bf{1) h(x) = f(x) + g(x)} \\\\:\implies \sf{h\left(\dfrac{-3}{2}\right) = f\left(\dfrac{-3}{2}\right) + g\left(\dfrac{-3}{2}\right)} \\\\:\implies \sf{h\left(\dfrac{-3}{2}\right) = \dfrac{3}{4} + \dfrac{-1}{4}} \\\\:\implies \sf{h\left(\dfrac{-3}{2}\right) = \dfrac{2}{4} = \dfrac{1}{2}}\\\\\text{As h(x) \neq 0, this isn't our answer}$

$\bf{2) p(x) = f(x) + 3g(x)} \\\\:\implies \sf{p\left(\dfrac{-3}{2}\right) = f\left(\dfrac{-3}{2}\right) + 3*g\left(\dfrac{-3}{2}\right)} \\\\:\implies \sf{p\left(\dfrac{-3}{2}\right) = \dfrac{3}{4} + 3*\dfrac{-1}{4}} \\\\:\implies \sf{p\left(\dfrac{-3}{2}\right) = \dfrac{3}{4} + \dfrac{-3}{4}} \\\\:\implies \sf{p\left(\dfrac{-3}{2}\right) = \dfrac{0}{4} = 0}\\\\\text{As p(x) = 0, this is our answer}$

$\bf{3) r(x) = 2f(x) + 3g(x)} \\\\:\implies \sf{r\left(\dfrac{-3}{2}\right) = 2*f\left(\dfrac{-3}{2}\right) + 3*g\left(\dfrac{-3}{2}\right)} \\\\:\implies \sf{r\left(\dfrac{-3}{2}\right) = 2*\dfrac{3}{4} + 3*\dfrac{-1}{4}} \\\\:\implies \sf{r\left(\dfrac{-3}{2}\right) = \dfrac{6}{4} + \dfrac{-3}{4}} \\\\:\implies \sf{r\left(\dfrac{-3}{2}\right) = \dfrac{3}{4}}\\\\\text{As r(x) \neq 0, this isn't our answer}$

$\bf{4) s(x) = 3f(x) + 2g(x)} \\\\:\implies \sf{s\left(\dfrac{-3}{2}\right) = 2*f\left(\dfrac{-3}{2}\right) + 3*g\left(\dfrac{-3}{2}\right)} \\\\:\implies \sf{s\left(\dfrac{-3}{2}\right) = 3*\dfrac{3}{4} + 2*\dfrac{-1}{4}} \\\\:\implies \sf{s\left(\dfrac{-3}{2}\right) = \dfrac{9}{4} + \dfrac{-2}{4}} \\\\:\implies \sf{s\left(\dfrac{-3}{2}\right) = \dfrac{7}{4}}\\\\\text{As s(x) \neq 0, this isn't our answer}$

Therefore, the polynomial which is divisible be 2x + 3, is

B) p(x) = f(x) + 3g(x).

Answer 2

Correct answer:-

• option B) p(x) = f(x) + 3g(x)

step-by-step solution:-

Put 2x + 3=0, we get x = − 3/2

Substitute x = − 3/2 in f(x) = 2x³+ 6x² + 4x as follows:

• f(x) = 2x³ + 6x² + 4x

→ f(-$\frac{3}{2}$ = 2(-$\frac{3}{2}$)³ + 6(-$\frac{3}{2}$)² + 4(-$\frac{3}{2}$)

→ f(-$\frac{3}{2}$) = 2(-$\frac{27}{8}$ + 6(-$\frac{9}{4}$ + 4(-$\frac{3}{2}$)

→ f$\frac{3}{4}$ = -$\frac{27}{4}$ + $\frac{54}{4}$ - 6

→ f(-$\frac{3}{2}$) = $\frac{27}{4}$ - 6

→ f(-$\frac{3}{2}$) = $\frac{3}{4}$

Now substitute x= −$\frac{3}{2}$ in g(x)

= x² + 3x+2 as follows:

g(x) = x² + 3x + 2

→ g(-$\frac{3}{2}$) = (-$\frac{3}{2}$)² + 3(-$\frac{3}{2}$) + 2

→ g(-$\frac{3}{2}$) = $\frac{9}{4}$ - $\frac{9}{2}$ + 2

→ g(-$\frac{3}{2}$) = $\frac{9-18+8}{4}$

→ g(-$\frac{3}{2}$) = -$\frac{1}{4}$)

Finally substituting x = − $\frac{3}{2}$ in the polynomial p(x) = f(x) + 3g(x) we get:

p(x)= f(x) + 3g(x)

→ p(-$\frac{3}{2}$) = f(-$\frac{3}{2}$) + 3g(-$\frac{3}{2}$)

→ p(-$\frac{3}{2}$) = f($\frac{3}{4}$) + 3(-$\frac{1}{4}$)

→ p(-$\frac{3}{2}$) = ($\frac{3}{4}$) - ($\frac{3}{4}$)

→ p(-$\frac{3}{2}$) = 0

Hence, the polynomial p(x) = f(x)+3g(x) is divisible by 2x+3.