By factor theorem, show that $(x+3)$ and $(2x-1)$ are factors of $2{{x}^{2}}+5x-3$ .

Hint: We need to show that $(x+3)$ and $(2x-1)$ are factors of $2{{x}^{2}}+5x-3$ by factor theorem. So first we will consider the given factors one by one. These are equated to 0. The value of $x$ so obtained will be substituted in the given polynomial. If the result of the simplification is equal to 0 in each case, then $(x+3)$ and $(2x-1)$ are factors of $2{{x}^{2}}+5x-3$ . Else they are not.

Complete step by step answer:
We need to show that $(x+3)$ and $(2x-1)$ are factors of $2{{x}^{2}}+5x-3$ .
For this, consider $(x+3)$ and equate it to 0.
i.e. $(x+3)=0$ .
Therefore, $x=-3$ .
Let $p(x)=2{{x}^{2}}+5x-3...(i)$
Substitute $x=-3$ in the above equation. We will get
$p(-3)=2{{(-3)}^{2}}+\left( 5\times -3 \right)-3$
By solving, we will get
$p(-3)=2\times 9-15-3$
After multiplication, the result will be
$\Rightarrow p(-3)=18-15-3$
$\Rightarrow p(-3)=0$
Thus $(x+3)$ is a factor of $2{{x}^{2}}+5x-3$ .
Now, consider $(2x-1)$ and equate it to 0.
i.e. $(2x-1)=0$
Collecting constant terms to one side, we will get
$2x=1$
Solving this, we get
$x=\dfrac{1}{2}$
Now, substitute this in equation $(i)$ . We will get
$p\left( \dfrac{1}{2} \right)=2{{\left( \dfrac{1}{2} \right)}^{2}}+5\left( \dfrac{1}{2} \right)-3$
By solving, we will get
$p\left( \dfrac{1}{2} \right)=2\times \dfrac{1}{4}+\dfrac{5}{2}-3$
$\Rightarrow p\left( \dfrac{1}{2} \right)=\dfrac{1}{2}+\dfrac{5}{2}-3$
Taking the LCM, we get
$p\left( \dfrac{1}{2} \right)=\dfrac{1+5-6}{2}$
By adding, we will get
$p\left( \dfrac{1}{2} \right)=0$
Thus $(2x-1)$ is a factor of $2{{x}^{2}}+5x-3$ .
Hence proved.




Note:
The factor theorem is used to obtain and verify the factors of a given polynomial. For this, the main part is the substitution of the exact value of the variable in the polynomial so that the polynomial will be equal to 0. Then only, we can say that the given data is a factor of the given polynomial.