Question

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
\[{t^2} - 3\], \[2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\]

Answer 1

Hint:
Here, we have to check whether the first polynomial is a factor of the first polynomial. If a polynomial is a factor of another polynomial, this means that the second polynomial is exactly divisible by the first polynomial, and leaves remainder 0 when divided by it. We will use a long division method to check whether the remainder is 0 or not, and hence, answer whether the first polynomial is a factor of the second polynomial or not.

Complete step by step solution:
We know that if \[g\left( x \right)\] is a factor of a polynomial \[p\left( x \right)\], then \[p\left( x \right)\] is exactly divisible by \[g\left( x \right)\].
This means that when \[p\left( x \right)\] is divided by \[g\left( x \right)\], the remainder should be equal to 0.
Therefore, we need to use a long division method to divide \[2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\] by \[{t^2} - 3\].
If the remainder is 0, then the polynomial \[{t^2} - 3\] is a factor of the polynomial \[2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\].
Now, let us use a long division method to divide \[2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\] by \[{t^2} - 3\].
Therefore, we get
\[\begin{array}{l}{t^2} - 3\mathop{\left){{1\begin{array}{l}2{t^4} + 3{t^3} - 2{t^2} - 9t - 12{\rm{ }}\\\underline {2{t^4}{\rm{ }} - 6{t^2}{\rm{ }}} \end{array}}}\right.
\!\!\!\!\overline{\,\,\ 1{\begin{array}{l}2{t^4} + 3{t^3} - 2{t^2} - 9t - 12{\rm{ }}\\\underline {2{t^4}{\rm{ }} - 6{t^2}{\rm{ }}} \end{array}}}}
\limits^{\displaystyle\,\,\, {2{t^2} + 3t + 4{\rm{ }}}}\\{\rm{ }}3{t^3} + 4{t^2} - 9t - 12\\{\rm{ }}\underline {3{t^3}{\rm{ }} - 9t{\rm{ }}} \\{\rm{ }}4{t^2}{\rm{ }} - 12{\rm{ }}\\{\rm{ }}\underline {4{t^2}{\rm{ }} - 12{\rm{ }}} {\rm{ }}\\{\rm{ }}\underline {{\rm{ 0 }}} {\rm{ }}\end{array}\]
We can observe that when \[2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\] is divided by \[{t^2} - 3\] using long division method, the remainder comes to be zero.
Therefore, \[{t^2} - 3\] is a factor of \[2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\].
We can also verify our answer by using the division algorithm.
The division algorithm states that if \[p\left( x \right)\] and \[g\left( x \right)\] are two polynomials where \[g\left( x \right) \ne 0\], then there are two polynomials \[q\left( x \right)\] and \[r\left( x \right)\] such that \[p\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)\]. In other words, \[{\rm{Dividend}} = {\rm{Quotient}} \times {\rm{Divisor}} + {\rm{Remainder}}\].
Substituting the dividend as \[2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\], quotient as \[2{t^2} + 3t + 4\], divisor as \[{t^2} - 3\], and remainder as 0, we get
\[ \Rightarrow 2{t^4} + 3{t^3} - 2{t^2} - 9t - 12 = \left( {2{t^2} + 3t + 4} \right)\left( {{t^2} - 3} \right) + 0\]
Multiplying the terms using distributive property, we get
\[ \Rightarrow 2{t^4} + 3{t^3} - 2{t^2} - 9t - 12 = 2{t^4} - 6{t^2} + 3{t^3} - 9t + 4{t^2} - 12 + 0\]
Adding and subtracting the like terms, we get
\[ \Rightarrow 2{t^4} + 3{t^3} - 2{t^2} - 9t - 12 = 2{t^4} + 3{t^3} - 2{t^2} - 9t - 12\]
Since LHS is equal to RHS, we have verified the answer.

Note:
We need to be careful while doing the long division. When multiplying \[{t^2} - 3\] by \[2{t^2}\], the product of \[ - 3\] and \[2{t^2}\] is \[ - 6{t^2}\], we have placed it under the term of the dividend that has the power of \[x\] as 2, instead of just next to \[2{t^4}\]. This helps in easy subtraction of the terms.
We have used the distributive law of multiplication in the solution to verify our answer. The distributive law of multiplication states that \[\left( {a + b + c} \right)\left( {d + e} \right) = a \cdot d + a \cdot e + b \cdot d + b \cdot e + c \cdot d + c \cdot e\].