(i) Find the value of \(a\) and of \(b\) for which \(2 x^2+3 x-2\) is a factor of \( 2 x^4+3 x^3+a\left(x^2+x\right)+b \)
(ii) Using the values of \(a\) and \(b\) found in part (i), solve the equation \(2 x^4+3 x^3+a\left(x^2+x\right)+b=0 \)
(i) If \(2 x^2+3 x-2\) is a factor of \( 2 x^4+3 x^3+a\left(x^2+x\right)+b \), the factors of \(2 x^2+3 x-2\) are also factors of \( 2 x^4+3 x^3+a\left(x^2+x\right)+b \)
Factorise \(2 x^2+3 x-2\) to get \( (2x-1)(x+2) \)
Apply factor theorem
Let \( f(x) = 2 x^4+3 x^3+a\left(x^2+x\right)+b \) ,
\( f(0.5) =0 \) , since \( (2x-1) \) is a factor
\( 2 + 3a +4b = 0 \)
\( f(-2) =0 \) , since \( (x+2) \) is a factor
\( 8 + 2a + b = 0 \)
Solving simultaneously, \(a = – 6 \), \( b = 4 \)
(ii) To solve \(2 x^4+3 x^3-6\left(x^2+x\right)+4=0 \)
Perform long division
The quotient of \( \left(2x^4 +3x^3 −6(x^2+x)+4 \right) \div \left(2x^2+3x−2 \right)\) is:\(x^2−2 \)
Hence \( \left(2x^2+3x−2 \right) \left(x^2−2 \right) = 0 \)