Question

What are the linear factors of the function f(x)=4x3−4x2−11x+6? Select each correct answer. (2x+3)
(2x−3)
(2x+1) (2x−1) (x−2) (x+2)
ps if you could can you please explain it...
Thank you so much I really appreciate it.

Answer 1

Step-by-step explanation:

Given:

$f(x) = 4 {x}^{3} - 4 {x}^{2} - 11x + 6$

To find:

What are the linear factors of the function

f(x)=4x³−4x²−11x+6

Select each correct answer. (2x+3)

(2x−3)

(2x+1) (2x−1) (x−2) (x+2)

Solution:

If any one of the linear polynomial is a factor of f(x) then by putting the value of x from the factor,converted f(x) into a zero.

Put value of x from every polynomial factor.

Case 1:

$2x + 3 = 0 \\ \\ 2x = - 3 \\ \\ x = \frac{ - 3}{2} \\ \\ f \bigg(\frac{ - 3}{2} \bigg) = 4 \bigg({\frac{ - 3}{2}}\bigg)^{3} - 4 \bigg({\frac{ - 3}{2}}\bigg)^{2} - 11\bigg(\frac{ - 3}{2} \bigg)+ 6 \\ \\ f \bigg(\frac{ - 3}{2} \bigg) = \frac{ - 27}{2} - 9 + \frac{33}{2} + 6 \\ \\ = \frac{ - 27 - 18 + 33 + 12}{2} \\ \\f \bigg(\frac{ - 3}{2} \bigg) = \frac{0}{2} \\ \\ f \bigg(\frac{ - 3}{2} \bigg) = 0$

Thus,

(2x+3) is a factor of polynomial f(x).

Case 2:

$2x - 3 = 0 \\ \\ 2x = 3 \\ \\ x = \frac{ 3}{2} \\ \\ f \bigg(\frac{ 3}{2} \bigg) = 4 \bigg({\frac{ 3}{2}}\bigg)^{3} - 4 \bigg({\frac{ 3}{2}}\bigg)^{2} - 11\bigg(\frac{ 3}{2} \bigg)+ 6 \\ \\ f \bigg(\frac{ 3}{2} \bigg) = \frac{ 27}{2} - 9 - \frac{33}{2} + 6 \\ \\ = \frac{ 27 - 18 - 33 + 12}{2} \\ \\f \bigg(\frac{3}{2} \bigg) = \frac{ - 12}{2} \\ \\ \\f \bigg(\frac{3}{2} \bigg) = - 6 \\ \\ f \bigg(\frac{3}{2} \bigg) \neq\:0$

This is not a factor of f(x).

Case 3:

$2x + 1 = 0 \\ \\ 2x = - 1 \\ \\ x = \frac{ - 1}{2} \\ \\ f \bigg(\frac{ - 1}{2} \bigg) = 4 \bigg({\frac{ - 1}{2}}\bigg)^{3} - 4 \bigg({\frac{ - 1}{2}}\bigg)^{2} - 11\bigg(\frac{ - 1}{2} \bigg)+ 6 \\ \\ f \bigg(\frac{ - 1}{2} \bigg) = \frac{ - 1}{2} - 1 + \frac{11}{2} + 6 \\ \\ = \frac{ - 1 - 2 + 11 + 12}{2} \\ \\f \bigg(\frac{ - 1}{2} \bigg) = \frac{20}{2} \\ \\ f \bigg(\frac{ - 1}{2} \bigg) = 10\\\\ f \bigg(\frac{ - 1}{2} \bigg) \neq 0$

This is not a factor of f(x).

Case 4:

$2x - 1 = 0 \\ \\ 2x = 1 \\ \\ x = \frac{ 1}{2} \\ \\ f \bigg(\frac{ - 1}{2} \bigg) = 4 \bigg({\frac{ 1}{2}}\bigg)^{3} - 4 \bigg({\frac{ 1}{2}}\bigg)^{2} - 11\bigg(\frac{ 1}{2} \bigg)+ 6 \\ \\ f \bigg(\frac{ 1}{2} \bigg) = \frac{ 1}{2} - 1 + \frac{11}{2} + 6 \\ \\ = \frac{ 1 - 2 - 11 + 12}{2} \\ \\f \bigg(\frac{ 1}{2} \bigg) = \frac{0}{2} \\ \\ f \bigg(\frac{ 1}{2} \bigg) = 0$

Thus,

(2x-1) is a factor of f(x).

Case 5:

$x - 2 = 0 \\ \\ x = 2 \\ \\ f(2) = 4 {(2)}^{3} - 4 {(2)}^{2} - 11(2) + 6 \\ \\ f(2) = 32 - 16 -22+ 6 \\ \\ f(2) = 0$

Thus,

(x-2) is a factor of f(x).

Case 6:

$x + 2 = 0 \\ \\ x = - 2 \\ \\ f(2) = 4 {( - 2)}^{3} - 4 {( - 2)}^{2} - 11( - 2) + 6 \\ \\ f( - 2) = - 32 - 16 + 22+ 6 \\ \\ f( - 2) = 20 \\ \\f( - 2)\:\neq\:0$

Thus,

(2x+3),(2x-1) and (x-2) are factors of polynomial f(x).

Hope it helps you.

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Answer 2

SOLUTION

TO DETERMINE

The linear factors of

$\sf{f(x) = 4 {x}^{3} - 4 {x}^{2} - 11x + 6 }$

EVALUATION

Here the given polynomial is

$\sf{f(x) = 4 {x}^{3} - 4 {x}^{2} - 11x + 6 }$

We now factorise it as below

$\sf{f(x) = 4 {x}^{3} - 4 {x}^{2} - 11x + 6 }$

For x = 2 we have f(2) = 0

∴ ( x - 2 ) is a factor of f(x)

$\sf{f(x) = 4 {x}^{3} - 4 {x}^{2} - 11x + 6 }$

$\sf{= 4 {x}^{3} -8 {x}^{2} + 4 {x}^{2} - 8x - 3x + 6 }$

$\sf{ = 4 {x}^{2} (x - 2) + 4x(x - 2) - 3(x - 2) }$

$\sf{ = (x - 2)(4 {x}^{2} + 4x - 3)}$

$\sf{ = (x - 2) \bigg[4 {x}^{2} +( 6 - 2)x - 3 \bigg] }$

$\sf{ = (x - 2) \bigg[4 {x}^{2} +6x - 2x- 3 \bigg] }$

$\sf{ = (x - 2) \bigg[2x(2x + 3) - 1(2x + 3) \bigg] }$

$\sf{ = (x - 2) \bigg[(2x + 3) (2x - 1) \bigg] }$

$\sf{ = (x - 2) (2x + 3) (2x - 1) }$

Hence the linear factors of

$\sf{f(x) = 4 {x}^{3} - 4 {x}^{2} - 11x + 6 } \: \: \: \: \: are$

$\sf{(x - 2) \: , \: (2x + 3) \: , \: (2x - 1)}$

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