Question

If \[{x^3} + a{x^2} - bx + 10\] is divided by \[{x^2} - 3x + 2\], find the value of \[a\] and \[b\].

Answer 1

Hint:
Here, we will use the concept of the factor theorem. We will factorize the given divisor to find two factors. We will then substitute the obtained factors in the given dividend to form two equations. We will subtract one equation from the other and solve it further to get the required values of \[a\] and \[b\].

Complete step by step solution:
It is given that \[{x^3} + a{x^2} - bx + 10\] is divided by \[{x^2} - 3x + 2\].
Let the polynomial be \[f\left( x \right) = {x^3} + a{x^2} - bx + 10\]……\[\left( 1 \right)\]
Its factor is \[g\left( x \right) = {x^2} - 3x + 2\]……..\[\left( 2 \right)\]
Now, we will factorize the equation \[\left( 2 \right)\] to get the two factors. Therefore, we get
\[\begin{array}{l}g\left( x \right) = {x^2} - 3x + 2\\ \Rightarrow g\left( x \right) = {x^2} - 2x - x + 2\end{array}\]
Taking \[x\] common in first two terms and \[ - 1\] common in last to terms, we get,
\[ \Rightarrow g\left( x \right) = x\left( {x - 2} \right) - 1\left( {x - 2} \right)\]
Factoring out the common terms, we get
\[ \Rightarrow g\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\].......\[\left( 3 \right)\]
It is given that equation \[\left( 1 \right)\] is divided by equation \[\left( 2 \right)\] and from equation \[\left( 3 \right)\] we know that our factors are 1 and 2.
So by Factor theorem \[f\left( 1 \right) = 0\] and\[f\left( 2 \right) = 0\].
So substituting \[x = 1\] in equation \[\left( 1 \right)\] we get,
\[f\left( 1 \right) = {1^3} + a \times {1^2} - b \times 1 + 10\]
Simplifying the expression, we get
\[ \Rightarrow f\left( 1 \right) = 1 + a - b + 10\]
Substituting \[f\left( 1 \right) = 0\] in above equation, we get
\[ \Rightarrow 0 = 1 + a - b + 10\]
Adding like terms, we get
\[ \Rightarrow a - b + 11 = 0\]……\[\left( 4 \right)\]
Now, substituting \[x = 2\] in equation \[\left( 2 \right)\] we get,
\[f\left( 2 \right) = {2^3} + a \times {2^2} - b \times 2 + 10\]
Simplifying the expression, we get
\[ \Rightarrow f\left( 2 \right) = 8 + 4a - 2b + 10\]
Substituting \[f\left( 2 \right) = 0\] in above equation, we get
\[ \Rightarrow 0 = 8 + 4a - 2b + 10\]
Adding like terms, we get
\[ \Rightarrow 4a - 2b + 18 = 0\]
Dividing both sides by 2, we get
\[ \Rightarrow 2a - b + 9 = 0\]……\[\left( 5 \right)\]
Next, we will subtract equation \[\left( 4 \right)\] from equation \[\left( 5 \right)\]. Therefore, we get
\[\begin{array}{l}\left( {2a - b + 9} \right) - \left( {a - b + 11} \right) = 0\\ \Rightarrow 2a - b + 9 - a + b - 11 = 0\end{array}\]
Adding and subtracting like terms, we get
\[\begin{array}{l} \Rightarrow a - 2 = 0\\ \Rightarrow a = 2\end{array}\]
Substituting value of \[a\] in equation \[\left( 5 \right)\], we get
\[2 \times 2 - b + 9 = 0\]
Simplifying the expression, we get

The value of \[a = 2\] and \[b = 13\].

Note:
Here we will find the value of a and b by using Factor theorem which state that if a polynomial \[f\left( x \right)\] of degree \[n \ge 1\] and ‘\[a\]’ is any real number then, if \[f\left( a \right) = 0\] then only \[\left( {x - a} \right)\] is a factor of the polynomial. The two problems where Factor Theorem is usually applied are when we have to factorize a polynomial and also when we have to find the roots of the polynomial. It is also used to remove known zeros from a polynomial while leaving all unknown zeros intact. It is a special case of a polynomial remainder theorem.