Question

The \[p(x)=2{{x}^{3}}-13{{x}^{2}}+23x-12\] divided by \[q(x)=2x-3\].

Answer 1

Hint:The given question is based on the topic “polynomials”. A polynomial \[p(x)\] in one variable \[x\] is an algebraic expression in x of the form \[p(x)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+...+{{a}_{2}}{{x}^{2}}+{{a}_{1}}x+{{a}_{0}}\] where \[{{a}_{0}},{{a}_{1}},{{a}_{2}},...,{{a}_{n}}\] are constants and \[{{a}_{n}}\ne 0\]. \[{{a}_{0}},{{a}_{1}},{{a}_{2}},...,{{a}_{n}}\] are respectively the coefficients of \[{{a}^{0}},a,{{a}^{1}},...,{{a}^{n}}\]. Here in this question we are going to use a long division method to divide the \[p(x)\] by \[q(x)\], where \[p(x)\] is the dividend and \[q(x)\] is the divisor.

Complete step by step answer:
Write the dividend and divisor in the long division format:
\[2{{x}^{3}}-13{{x}^{2}}+23x-12\] is divided by \[2x-3\]
We need to find the first term of the quotient for this, divide the first term of the dividend \[2{{x}^{3}}\] by the first term of the divisor \[2x\], \[\dfrac{2{{x}^{3}}}{2x}\], \[2\] and \[x\] will be get cancel and the answer is \[{{x}^{2}}\].

Now in long division method use \[{{x}^{2}}\], multiply the divisor by the first term of the quotient, and subtract this product from the dividend, that is, \[{{x}^{2}}(2x-3)=2{{x}^{3}}-3{{x}^{2}}\], subtract it from the dividend,
\[2x-3\overset{{x}^{2}}{\overline{\left){\begin{align}
  & 2x^{3}-13x^{2}+23x-12 \\
 & \underline{2x^{3}-3x^{2}} \\
 & 0\;\;\;\;-10x^{2}+23x-12 \\
\end{align}}\right.}}\]

Now, we treat \[-10{{x}^{2}}+23x-12\] as the new dividend. To find the second term of the quotient, we need to divide the first term of the new dividend \[-10{{x}^{2}}\] by the first term of the divisor \[2x\], \[\dfrac{-10{{x}^{2}}}{2x}=-5x\].Now multiply the divisor by the second term of the quotient, and subtract this product from the dividend, i.e., \[-5x(2x-3)=-10{{x}^{2}}+15x\], subtract it from the dividend,
\[2x-3\overset{{x}^{2}-5x}{\overline{\left){\begin{align}
  & 2x^{3}-13x^{2}+23x-12 \\
 & \underline{2x^{3}-3x^{2}} \\
 & \;\;\;\;-10x^{2}+23x-12 \\
& \underline{\;\;\;\;\;-10x^{2}+15x}\\
& \,\,\,\,\,\,\,\,\,\,\,{8x-12}
\end{align}}\right.}}\]

Now our new dividend is \[8x-12\], to find the quotient, divide the first term of the dividend \[8x\] by the first term of the divisor \[2x\], \[\dfrac{8x}{2x}=4\].
By multiplying \[(2x-3)4=8x-12\], applying in long division,
\[2x-3\overset{{x}^{2}-5x+4}{\overline{\left){\begin{align}
  & 2x^{3}-13x^{2}+23x-12 \\
 & \underline{2x^{3}-3x^{2}} \\
 & \;\;\;\;-10x^{2}+23x-12 \\
& \underline{\;\;\;\;\;-10x^{2}+15x}\\
& \,\,\,\,\,\,\,\,\,\,\,{8x-12}\\
& \,\,\,\,\,\,\,\,\,\,\,\underline{8x-12}\\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0
\end{align}}\right.}}\]
Here the remainder gets \[0\], so further cannot be divided.

Hence the quotient of \[2{{x}^{3}}-13{{x}^{2}}+23x-12\] divided by \[2x-3\] is \[{{x}^{2}}-5x+4\] and remainder is \[0\].

Note:We can see that \[2{{x}^{3}}-13{{x}^{2}}+23x-12=(2x-3)({{x}^{2}}-5x+4)+0\], that is, \[\text{Dividend=}\left( \text{Divisor}\times \text{Quotient} \right)\text{+Remainder}\].In general, if \[p(x)\] and \[g(x)\] are two polynomials such that degree of \[p(x)\ge \] degree of \[g(x)\] and \[g(x)\ne 0\]then we can find polynomials \[q(x)\] and \[r(x)\] such that: \[p(x)=g(x)q(x)+r(x)\], \[r(x)=0\]or degree of \[r(x)<\] degree of \[g(x)\]. Here we say that \[p(x)\] divided by \[g(x)\] gives \[q(x)\] as quotient and \[r(x)\] as remainder.