Question

Given that ${x^2} + x - 6$ is a factor of $2{x^4} + {x^3} - a{x^2} + bx + a + b - 1$, then find the values of $a$ and $b$.

A. $a = 16,b = 3$

B. $a = 16,b = 2$

C. $a = 10,b = 3$

D. None of these

Answer 1

Hint- In order to deal with this question first we will get the $2$ roots of the given equation by equal the given factor to zero further we will put the obtained root in the given equation and get $2$ equations with $2$ variables and by solving it we will get the required answer.

Complete step-by-step answer:
Given equation is $2{x^4} + {x^3} - a{x^2} + bx + a + b - 1$
Now in order to find the roots we will equal the given factor to zero

$
  {x^2} + x - 6 = (x + 3)(x - 2) \\
   \Rightarrow (x + 3)(x - 2) = 0 \\
   \Rightarrow (x + 3) = 0{\text{ and }}(x - 2) = 0 \\
   \Rightarrow x = - 3{\text{ and }}x = 2 \\
$
So the roots of given equation is $2$ and \[ - 3\]
As we know that roots of equation always satisfy the corresponding equation
Therefore, obtained roots $2$ and \[ - 3\] also satisfy the corresponding equation

So, \[f\left( { - 3} \right){\text{ }} = {\text{ }}0{\text{ }}and{\text{ }}f\left( 2 \right){\text{ }} = {\text{ }}0\]

By considering \[f( - 3) = 0\], we get

\[
  f( - 3) = 2{( - 3)^4} + {( - 3)^3} - a{( - 3)^2} + b( - 3) + a + b - 1 \\
   \Rightarrow 2{( - 3)^4} + {( - 3)^3} - a{( - 3)^2} + b( - 3) + a + b - 1 = 0 \\
   \Rightarrow 134 - 8a - 2b = 0 \\
   \Rightarrow 4a + b = 67 \to (1) \\
\]
By taking \[f\left( 2 \right){\text{ }} = {\text{ }}0\], we have

\[
  f\left( 2 \right){\text{ }} = {\text{ 2(2}}{{\text{)}}^4} + {2^3} - a{(2)^2} + 2b + a + b - 1 \\
   \Rightarrow {\text{2(2}}{{\text{)}}^4} + {2^3} - a{(2)^2} + 2b + a + b - 1 = 0 \\
   \Rightarrow 39 - 3a + 3b = 0 \\
   \Rightarrow a - b = 13 \to (2) \\
\]
Now we get two equations with $2$ variables
\[
  4a + b = 67 \to (1) \\
  a - b = 13 \to (2) \\
\]
  We will solve it by substitution method so we will put the value of \[b\] from equation \[1\] to equation \[2\]
\[b = a - 13\] from equation \[2\]
\[
  4a + (a - 13) = 67 \\
  5a - 13 = 67 \\
  5a = 80 \\
   \Rightarrow a = 16 \\
\]
Therefore we get the value of \[a\] as \[16\]
Now substitute value of \[a\] in equation \[2\] we get
\[
  a - b = 13 \\
  16 - b = 13 \\
   \Rightarrow b = 16 - 13 = 3 \\
\]

Hence the values of \[a\] and \[b\] are \[16\] and \[3\] respectively and the correct answer is option A.

Note- In mathematics, the breaking apart of a polynomial into a series of other smaller polynomials is factorization or factoring. You can then add these factors together if you like, and you will get the initial polynomial. For example $(x + 1)$and $(x - 1)$is a factor of $({x^2} - 1)$. By multiplying these factors we get the original polynomial so we can say that $({x^2} - 1) = (x - 1)(x + 1)$.