Answer

Verified

340.5k+ views

**Hint:**We factor the given equation with the help of vanishing method. In this method we find a number $a$ such that for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We assume $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24$ and take the value of $a$ as 4.

**Complete step by step solution:**

We find the value of $x=a$ for which the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24=0$.

We take $x=a=4$.

We can see $f\left( 4 \right)={{4}^{3}}+{{4}^{2}}-14\times 4-24=64+16-56-24=0$.

So, the root of the $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24$ will be the function $\left( x-4

\right)$.

This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.

Therefore, the term $\left( x-4 \right)$ is a factor of the polynomial ${{x}^{3}}+{{x}^{2}}-14x-24$.

We can now divide the polynomial ${{x}^{3}}+{{x}^{2}}-14x-24$ by $\left( x-4 \right)$.

\[x-4\overset{{{x}^{2}}+5x+6}{\overline{\left){\begin{align}

& {{x}^{3}}+{{x}^{2}}-14x-24 \\

& \underline{{{x}^{3}}-4{{x}^{2}}} \\

& 5{{x}^{2}}-14x \\

& \underline{5{{x}^{2}}-20x} \\

& 6x-24 \\

& \underline{6x-24} \\

& 0 \\

\end{align}}\right.}}\]

We first tried to equate the highest power of the dividend with the highest power of the divisor and that’s why we multiplied with ${{x}^{2}}$. We get \[{{x}^{3}}-4{{x}^{2}}\]. We subtract it to get

\[5{{x}^{2}}-14x\]. We again equate with the highest power of the remaining terms. We multiply with

$5x$ and subtract to get \[6x-24\]. At the end we had to multiply with 6 to complete the division.

The quotient is \[{{x}^{2}}+5x+6\].

We still can factor \[{{x}^{2}}+5x+6\].

We apply a grouping method where \[{{x}^{2}}+5x+6={{x}^{2}}+3x+2x+6\].

\[\begin{align}

& {{x}^{2}}+3x+2x+6 \\

& =x\left( x+3 \right)+2\left( x+3 \right) \\

& =\left( x+3 \right)\left( x+2 \right) \\

\end{align}\]

**Therefore, the factored form of \[{{x}^{2}}+5x+6\] is \[\left( x+3 \right)\left( x+2 \right)\].**

The final factorisation is ${{x}^{3}}+{{x}^{2}}-14x-24=\left( x-4 \right)\left( x+3 \right)\left( x+2

\right)$.

The final factorisation is ${{x}^{3}}+{{x}^{2}}-14x-24=\left( x-4 \right)\left( x+3 \right)\left( x+2

\right)$.

**Note:**We find the value of x for which the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24=0$.

We can see $f\left( -2 \right)={{\left( -2 \right)}^{3}}+{{\left( -2 \right)}^{2}}-14\times \left( -2 \right)-24=-8+4+28-24=0$. So, the root of the $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24$ will be the function $\left( x+2 \right)$. We can also do this for $\left( x+3 \right)$.

Recently Updated Pages

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

What happens when entropy reaches maximum class 11 chemistry JEE_Main

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Can anyone list 10 advantages and disadvantages of friction

State and prove Bernoullis theorem class 11 physics CBSE

The ice floats on water because A solid have lesser class 9 chemistry CBSE

State Newtons formula for the velocity of sound in class 11 physics CBSE