By division algorithm, when p(x) = 3x3 + x2 − 22x + 9 is divided by `3x^2 + 7x - 6,`the reminder is a linear polynomial. So, let r(x) = ax + b be added to p(x) so that the result is divisible by q(x) Let `f(x) = p(x) + r(x)` ` = 3x^2 + x^2 - 22x + 9 ax +b` ` = 3x^2 + x^2 +(a- 22) x + 9 + b` We have \[q\left( x \right) = 3 x^2 + 7x - 6\] \[ = 3 x^2 + 9x - 2x - 6\] \[ = 3x\left( x + 3 \right) - 2\left( x + 3 \right)\] \[ = \left( 3x - 2 \right) \left( x + 3 \right)\] Clearly,
\[\left( 3x - 2 \right)\] and \[\left( x + 3 \right)\]
are factors of q(x). Therefore, f(x) will be divisible by q(x) if (3x - 2)and (x + 3)are factors of f(x), i.e., `f (2/3)`and f(−3) are equal to zero. Now, \[f\left( \frac{2}{3} \right) = 0\] \[ \Rightarrow 3 \left( \frac{2}{3} \right)^3 + \left( \frac{2}{3} \right)^2 + \left( a - 22 \right)\left( \frac{2}{3} \right) + 9 + b = 0\] \[ \Rightarrow 3 \times \frac{8}{27} + \frac{4}{9} + \frac{2a}{3} - \frac{44}{3} + 9 + b = 0\] \[ \Rightarrow \frac{8}{9} + \frac{4}{9} - \frac{44}{3} + 9 + \frac{2a}{3} + b = 0\] \[ \Rightarrow \frac{8 + 4 - 132 + 81}{9} + \frac{2a}{3} + b = 0\] \[ \Rightarrow - \frac{39}{9} + \frac{2a}{3} + b = 0\] \[ \Rightarrow \frac{2a}{3} + b = \frac{13}{3}\] \[ \Rightarrow 2a + 3b = 13 . . . . . . . . \left( i \right)\] And \[f\left( - 3 \right) = 0\] \[ \Rightarrow 3 \left( - 3 \right)^3 + \left( - 3 \right)^2 + \left( a - 22 \right)\left( - 3 \right) + 9 + b = 0\] \[ \Rightarrow - 81 + 9 - 3a + 66 + 9 + b = 0\] \[ \Rightarrow - 3a + b = - 3 \] \[ \Rightarrow b = - 3 + 3a . . . . . . . . . \left( ii \right)\] Substituting the value of b from (ii) in (i), we get, \[2a + 3\left( 3a - 3 \right) = 13\] \[ \Rightarrow 2a + 9a - 9 = 13\] \[ \Rightarrow 11a = 13 + 9\] \[ \Rightarrow 11a = 22\] \[ \Rightarrow a = 2\] Now, from (ii), we get \[b = - 3 + 3\left( 2 \right) = - 3 + 6 = 3\] So, we have a = 2 and b = 3 Hence, p(x) is divisible by q(x), if 2x + 3is added to it. |