
Find the value of a and b so the polynomial \[{{\text{x}}^{\text{4}}}{\text{ + a}}{{\text{x}}^{\text{3}}}{\text{ - 7}}{{\text{x}}^{\text{2}}}{\text{ - 8x + b}}\] exactly divisible by \[{\text{(x + 2)}}\] as well \[{\text{(x + 3)}}\].
Answer
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Hint: As the factors are given of the polynomial or we can say zeroes of the polynomial are given. So, put the zeroes of the polynomial in the given polynomial and form two-equation from it and solve two equations in two variable methods to find the value of a and b.
Complete step by step answer:
Let the given polynomial be \[{\text{P(x) = 0}}\] and so ,
\[{{\text{x}}^{\text{4}}}{\text{ + a}}{{\text{x}}^{\text{3}}}{\text{ - 7}}{{\text{x}}^{\text{2}}}{\text{ - 8x + b = 0}}\]
And now , as given its factors are , \[{\text{(x + 2) and (x + 3)}}\]
Here calculating the value of x as the zeroes of the polynomial , so the values of x are
\[{\text{(x = - 2),(x = - 3)}}\]
Now substitute the values of x in the given polynomial,
\[
{\text{P( - 2) = ( - 2}}{{\text{)}}^{\text{4}}}{\text{ + a( - 2}}{{\text{)}}^{\text{3}}}{\text{ - 7( - 2}}{{\text{)}}^{\text{2}}}{\text{ - 8( - 2) + b = 0}} \\
\Rightarrow {\text{16 - 8a - 28 + 16 + b = 0}} \\
\Rightarrow {\text{4 + b = 8a}} \\
\\
{\text{P( - 3) = ( - 3}}{{\text{)}}^{\text{4}}}{\text{ + a( - 3}}{{\text{)}}^{\text{3}}}{\text{ - 7( - 3}}{{\text{)}}^{\text{2}}}{\text{ - 8( - 3) + b = 0}} \\
\Rightarrow {\text{81 - 27a - 63 + 24 + b = 0}} \\
\Rightarrow {\text{27a = b + 42}} \\
\]
Now , substituting value of b from \[{\text{4 + b = 8a}}\] to \[{\text{27a = b + 42}}\], we get,
\[
{\text{27a = b + 42}} \\
\Rightarrow {\text{27a = 8a - 4 + 42}} \\
\Rightarrow {\text{19a = 38}} \\
\Rightarrow {\text{a = 2}} \\
\]
Now substitute the value of a in any one of the equations and calculate the value of b.
\[
{\text{4 + b = 8a}} \\
{\text{as,a = 2}} \\
\Rightarrow {\text{b = 8(2) - 4}} \\
\Rightarrow {\text{b = 16 - 4}} \\
\Rightarrow {\text{b = 12}} \\
\]
Thus the values of a and b are \[{\text{(2,12)}}\].
Note: We know that a polynomial is an algebraic term with one or many terms. Zeros of polynomials are the real values of the variable for which the value of the polynomial becomes zero. So, real numbers, 'm' and 'n' are zeros of polynomial p(x), if p(m) \[ = 0\] and p(n) \[ = 0\]
Complete step by step answer:
Let the given polynomial be \[{\text{P(x) = 0}}\] and so ,
\[{{\text{x}}^{\text{4}}}{\text{ + a}}{{\text{x}}^{\text{3}}}{\text{ - 7}}{{\text{x}}^{\text{2}}}{\text{ - 8x + b = 0}}\]
And now , as given its factors are , \[{\text{(x + 2) and (x + 3)}}\]
Here calculating the value of x as the zeroes of the polynomial , so the values of x are
\[{\text{(x = - 2),(x = - 3)}}\]
Now substitute the values of x in the given polynomial,
\[
{\text{P( - 2) = ( - 2}}{{\text{)}}^{\text{4}}}{\text{ + a( - 2}}{{\text{)}}^{\text{3}}}{\text{ - 7( - 2}}{{\text{)}}^{\text{2}}}{\text{ - 8( - 2) + b = 0}} \\
\Rightarrow {\text{16 - 8a - 28 + 16 + b = 0}} \\
\Rightarrow {\text{4 + b = 8a}} \\
\\
{\text{P( - 3) = ( - 3}}{{\text{)}}^{\text{4}}}{\text{ + a( - 3}}{{\text{)}}^{\text{3}}}{\text{ - 7( - 3}}{{\text{)}}^{\text{2}}}{\text{ - 8( - 3) + b = 0}} \\
\Rightarrow {\text{81 - 27a - 63 + 24 + b = 0}} \\
\Rightarrow {\text{27a = b + 42}} \\
\]
Now , substituting value of b from \[{\text{4 + b = 8a}}\] to \[{\text{27a = b + 42}}\], we get,
\[
{\text{27a = b + 42}} \\
\Rightarrow {\text{27a = 8a - 4 + 42}} \\
\Rightarrow {\text{19a = 38}} \\
\Rightarrow {\text{a = 2}} \\
\]
Now substitute the value of a in any one of the equations and calculate the value of b.
\[
{\text{4 + b = 8a}} \\
{\text{as,a = 2}} \\
\Rightarrow {\text{b = 8(2) - 4}} \\
\Rightarrow {\text{b = 16 - 4}} \\
\Rightarrow {\text{b = 12}} \\
\]
Thus the values of a and b are \[{\text{(2,12)}}\].
Note: We know that a polynomial is an algebraic term with one or many terms. Zeros of polynomials are the real values of the variable for which the value of the polynomial becomes zero. So, real numbers, 'm' and 'n' are zeros of polynomial p(x), if p(m) \[ = 0\] and p(n) \[ = 0\]
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