Find the values of a and b when the polynomials f\left( x \right)=2{{x}^{2}}-5x+aandg\left( x \right)=2{{x}^{2}}+5x+b both have a factor \left( 2x+1 \right)
Find the values of a and b when the polynomials f\left( x \right)=2{{x}^{2}}-5x+aandg\left( x \right)=2{{x}^{2}}+5x+b both have a factor \left( 2x+1 \right)

Remainder theorem is a theorem in algebra: if f(x) is a polynomial in x then the remainder on dividing f(x) by x-a is f(a).

In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial. It is special case of the polynomial remainder theorem.

Solution:

From the question it is given that,

f\left( x \right)=2{{x}^{2}}-5x+a

g\left( x \right)=2{{x}^{2}}+5x+b

Let us assume 2x+1=0,x=-{}^{1}/{}_{2}

Now, substitute the value of x in f\left( x \right) we get,

f\left( {{-}^{1}}{{/}_{2}} \right)=2{{\left( {{-}^{1}}{{/}_{2}} \right)}^{2}}-5\left( {{-}^{1}}{{/}_{2}} \right)+a=0

2\left( 1/4 \right)+5/2+a=0

^{1}{{/}_{2}}+5/2+a=0

6/2+a=0

3+a=0

a=-3

Then,

g\left( {{-}^{1}}{{/}_{2}} \right)=2{{\left( {{-}^{1}}{{/}_{2}} \right)}^{2}}+5\left( {{-}^{1}}{{/}_{2}} \right)+b=0

2\left( 1/4 \right)-5/2+b=0

^{1}{{/}_{2}}-5/2+b=0

-4/2+b=0

-2+b=0

b=2

Therefore, the value of a is -3and b  is 2.