Question

Nonsense/No solution = report

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Answer 1

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

Given cubic polynomial is

$\rm \: {4x}^{3} - {7x}^{2} - 14x - 3$

Let assume that

$\rm \: f(x) = {4x}^{3} - {7x}^{2} - 14x - 3$

First we have to find the zero of f(x) by using hit and trial method.

So, Let assume that x = - 1

$\rm \: f( - 1) = {4( - 1)}^{3} - {7( - 1)}^{2} - 14( - 1) - 3$

$\rm \: f( - 1) = - 4 - 7 + 14 - 3$

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

We know,

Factor theorem states that if a polynomial f(x) of degree more than 1 is such that f(- a) = 0, then x + a is factor of f(x).

So, By factor theorem,

$\rm\implies \:x + 1 \: is \: a \: factor \: of \: f(x)$

Now, By using Method of Synthetic Division, we have

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{{\sf{\:\: \:\:}}}\\ {\underline{\sf{ - 1}}}& {\sf{\: 4 \: \: \: \: \: - 7 \: \: \: \: \: - 14 \: \: \: \: - 3 \:\:}} \\{\sf{}}& \underline{\sf{\:\: \:  \:  \:  - 4 \: \: \: \: \: \: \: \: \: \: 11\:  \:  \:  \:  \: 3\: \:\:}} \\ {{\sf{}}}& {\sf{\: \:  \:  \:  \: 4 \:  \:  \:  \:  - 11\:  \:  \:  - 3  \:  \:  \:  [ \: 0\: \: \:  \:  \:  \:   \:\:}} \\ {\sf{\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$

Now, By using Division Algorithm, we have

Dividend = Divisor × Quotient + Remainder

So,

$\rm \: {4x}^{3} - {7x}^{2} - 14x - 3$

$\rm \: = \: (x + 1)( {4x}^{2} - 11x - 3)$

$\rm \: = \: (x + 1)( {4x}^{2} - 12x + x - 3)$

$\rm \: = \: (x + 1)\bigg[4x(x - 3) + 1(x - 3)\bigg]$

$\rm \: = \: (x + 1)\bigg[4x(x - 3) + 1(x - 3)\bigg]$

$\rm \: = \: (x + 1)(x - 3)(4x + 1)$

Hence,

$\boxed{\tt{ \rm \: {4x}^{3} - {7x}^{2} - 14x - 3 = (x + 1)(x - 3)(4x + 1)}}$

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)