Question

The function is defined by f(x)= 2xcube + 7xsquare+7x+2, find the linear factors of f(x) and sketch the curve of fx

Answer 1

$\large\underline{\sf{Given- }}$

The function,

$\rm :\longmapsto\:f(x) = {2x}^{3} + {7x}^{2} + 7x + 2$

$\large\underline{\sf{To\:Find - }}$

1. The linear factors of f(x)

2. The graph of f(x).

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {2x}^{3} + {7x}^{2} + 7x + 2$

can be regrouped as

$\rm \: = \: \: ( {2x}^{3} + 2) + ( {7x}^{2} + 7x)$

$\rm \: = \: \: 2( {x}^{3} + 1) + 7( {x}^{2} + x)$

We know,

$\boxed{ \rm{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2})}}$

So, using this identity, we get

$\rm \: = \: \: 2(x + 1)( {x}^{2} - x + 1) + 7x(x + 1)$

On taking ( x + 1 ) common, we get

$\rm \: = \: \: (x + 1)(2 {x}^{2} - 2x + 2 + 7x)$

$\rm \: = \: \: (x + 1)(2 {x}^{2} + 5x + 2)$

$\rm \: = \: \: (x + 1)(2 {x}^{2} + 4x + x + 2)$

$\rm \: = \: \: (x + 1)\bigg(2x(x + 2) + 1(x + 2)\bigg)$

$\rm \: = \: \: (x + 1)(x + 2)(2x + 1)$

Hence,

$\boxed{ \bf \: { f(x) \: = \: (x + 1)(x + 2)(2x + 1) \: }}$

To sketch the graph of f(x).

Let assume that

$\boxed{ \bf \: { y \: = \: f(x) \: = \: (x + 1)(x + 2)(2x + 1) \: }}$

Point of intersection with x - axis.

On x - axis, y = 0

$\rm :\longmapsto\:(x + 1)(x + 2)(2x + 1) = 0$

$\bf\implies \:x = - 1, \: - 2, \: - \: \dfrac{1}{2}$

Hence, Point of intersection with x- axis are

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf - 1 & \sf 0 \\ \\ \sf - 2 & \sf 0 \\ \\ \sf - \dfrac{1}{2} & \sf 0 \end{array}} \\ \end{gathered}$

Point of intersection with y - axis

On y - axis, x = 0

$\rm :\longmapsto\:y = (0 + 1)(0 + 2)(2(0) + 1)$

$\rm :\longmapsto\:y = 1 \times 2 \times 1$

$\bf\implies \:y = 2$

So, point of intersection with y axis is ( 0, 2 ).

➢ Now draw a graph using the points

➢ See the attachment graph.