Question

Help me with this, this is my last question this day . pls don't spam.​

Answer 1

$\large\underline{\sf{Solution-3}}$

Given polynomial is

$\rm \: {x}^{3} + {4x}^{2} + x - 6$

Let assume that

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

Let check x = - 3

$\rm \: f( - 3) = {( - 3)}^{3} + {4( - 3)}^{2} - 3 - 6$

$\rm \: f( - 3) = - 27 + 36 - 9$

$\rm \: f( - 3) = 0$

$\rm\implies \: - \: 3 \: is \: its \: solution$

Now, Let check - 2

$\rm \: f( - 2) = {( - 2)}^{3} + {4( - 2)}^{2} - 2 - 6$

$\rm \: f( - 2) = - 8 + 16 - 2 - 6$

$\rm \: f( - 2) = 0$

$\rm\implies \: - \: 2 \: is \: its \: solution$

Now, Let's check x = 1

$\rm \: f(1) = 1 + 4 + 1 - 6$

$\rm \: f(1) = 0$

$\rm\implies \: 1\: is \: its \: solution$

Let's check x = 2

$\rm \: f(2) = {(2)}^{3} + {4(2)}^{2} + 2 - 6$

$\rm \: f(2) = 8 + 16 - 4$

$\rm \: f(2) = 20$

$\rm\implies \: 2\: is \: not\: its \: solution$

So, Option D. is correct.

$\large\underline{\sf{Solution-4}}$

Given polynomial is

$\rm \: {x}^{4} + {5x}^{3} + {6x}^{2} - 4x - 8$

So, By synthetic Division, we have

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{{\sf{\:\: \:\:}}}\\ {\underline{\sf{ - 2}}}& {\sf{\: 1 \: \: \: \: \: 5 \: \: \: \: \: \: \: 6 \: \: \: \: - 4 \: \: \: \: \: \: - 8 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \: \: \: - 2 \: \: - 6\:\:  \:  \:  \: \: 0\: \:  \: \: \: - 8 \: \: }} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  1\:  \:  \:  3\:  \:  \: 0  \:  \:  \:  - 4 \: \: \: \: \: \: \: [ \: 0\: \: \:  \:  \:  \:   \:\:}} \\ {\sf{\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$

So, x + 2 is its factor

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{{\sf{\:\: \:\:}}}\\ {\underline{\sf{ - 2}}}& {\sf{\: 1 \: \: \: \: \: 3\: \: \: \: \: \: \: 0 \: \: \: \: - 4 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \: - 2 \: \: \: - 2\:  \:  \:  \:  \: 4\: \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \: \: \: \: 1\:  \:  \:  1\:  - 2  \:  \:  \:  [ \: 0\: \: \:  \:  \:  \:   \:\:}} \\ {\sf{\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$

So, x + 2 is its factor

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{{\sf{\:\: \:\:}}}\\ {\underline{\sf{ - 2}}}& {\sf{\: 1 \: \: \: \: \: 1\: \: \: \: \: - 2 }} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  - 2\: \: \: \: \: \: \: \: \: 2\:  \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \:1  \:  - 1\:    \:  \: \: [ \: 0\: \: \:  \:  \:  \:   \:\:}} \\ {\sf{\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$

So, x + 2 is its factor

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{{\sf{\:\: \:\:}}}\\ {\underline{\sf{ - 2}}}& {\sf{\: 1 \: \: \: \: \: - 1\: }} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  \: \: \: \: \: \: \: - 2\: \: \: \: \:   \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \:  \:  \: \: \: \: \: 1  \: \: \: \: \: [ \: - 3\: \: \:  \:  \:  \:   \:\:}} \\ {\sf{\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$

So, x + 2 is not a factor

So, x + 2 repeat three times

So, option C is correct

$\large\underline{\sf{Solution-5}}$

Let assume that

$\rm \: a = - 2$

$\rm \: b = 3$

$\rm \: c = 5$

$\rm \: S_1 = a + b + c$

$\rm \: = \: - 2 + 3 + 5$

$\rm \: = \: 6$

$\rm\implies \:S_1 = 6$

Now, Consider

$\rm \: S_2 = ab + bc + ca$

$\rm \: = \: - 2 \times 3 + 3 \times 5 + 5 \times ( - 2)$

$\rm \: = \: - 6 + 15 - 10$

$\rm \: = \: - 1$

$\rm\implies \:S_2 \: = \: - \: 1$

Now, Consider

$\rm \: S_3 = abc$

$\rm \: = \: - 2 \times 3 \times 5$

$\rm \: = \: - 30$

$\rm\implies \:S_3 \: = \: - \: 30$

So, The required polynomial is

$\rm \: f(x) = {x}^{3} - S_1 {x}^{2} + S_2x - S_3$

$\rm \: f(x) = {x}^{3} - 6{x}^{2} - x + 30$

So, Option B is correct.

Answer 2

Answer:

solution 3 ans is b as above explanation is given

Thanx