Answer

Verified

375k+ views

**Hint:**We use both grouping method and vanishing method to find the factor of the problem. We take common terms out to form the multiplied forms. In the case of the vanishing method, we use the value of x which gives the polynomial value 0.

**Complete step-by-step solution:**

We apply the middle-term factoring or grouping to factorise the polynomial.

Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number.

In the case of $2{{x}^{2}}-11x+5$, we break the middle term $-11x$ into two parts of $-10x$ and $-x$.

So, $2{{x}^{2}}-11x+5=2{{x}^{2}}-10x-x+5$. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.

Here multiplication for both cases gives $10{{x}^{2}}$. The grouping will be done for $2{{x}^{2}}-10x$ and $-x+5$. We try to take the common numbers out.

For $2{{x}^{2}}-10x$, we take $2x$ and get $2x\left( x-5 \right)$.

For $-x+5$, we take $-1$ and get $-\left( x-5 \right)$.

The equation becomes $2{{x}^{2}}-11x+5=2{{x}^{2}}-10x-x+5=2x\left( x-5 \right)-\left( x-5 \right)$.

Both the terms have $\left( x-5 \right)$ in common. We take that term again and get

$\begin{align}

& 2{{x}^{2}}-11x+5 \\

& =2x\left( x-5 \right)-\left( x-5 \right) \\

& =\left( x-5 \right)\left( 2x-1 \right) \\

\end{align}$

**Therefore, the factorisation of $2{{x}^{2}}-11x+5$ is $\left( x-5 \right)\left( 2x-1 \right)$.**

**Note:**We find the value of x for which the function $f\left( x \right)=2{{x}^{2}}-11x+5=0$. We can see $f\left( 5 \right)=2\times {{5}^{2}}-11\times 5+5=50-55+5=0$. So, the root of the $f\left( x \right)=2{{x}^{2}}-11x+5$ will be the function $\left( x-5 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.

Now, $f\left( x \right)=2{{x}^{2}}-11x+5=\left( x-5 \right)\left( 2x-1 \right)$. We can also do this for $\left( 2x-1 \right)$.

Recently Updated Pages

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

The provincial president of the constituent assembly class 11 social science CBSE

Gersoppa waterfall is located in AGuyana BUganda C class 9 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

The hundru falls is in A Chota Nagpur Plateau B Calcutta class 8 social science CBSE

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

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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