Question

Prove that \[(3x + 7)\] is a factor of \[(6{x^3} + 29{x^2} + 44x + 21)\]

Answer 1

Hint:The factor of any polynomial expression will be an expression multiplied by a number or another expression. Here we are asked to prove that the given factor is a factor of the given expression. This can be done by checking it by dividing the given expression by the given factor, if we get the remainder as zero then it is a factor of that expression otherwise it is not.

Complete step-by-step solution:
We aim to verify that \[(3x + 7)\] is the factor of the expression \[(6{x^3} + 29{x^2} + 44x + 21)\].
We can check this by dividing the expression \[(6{x^3} + 29{x^2} + 44x + 21)\] by the factor \[(3x + 7)\].
Let us start dividing the expression. When the expression \[(3x + 7)\] is multiplied by \[2{x^2}\]we get \[6{x^3} + 14{x^2}\] on subtracting this from the dividend we will easily eliminate the first term from the dividend.
\[3x + 7\mathop{\left){\vphantom{1
  6{x^3} + 29{x^2} + 44x + 21 \\
  6{x^3} + 14{x^2} \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  6{x^3} + 29{x^2} + 44x + 21 \\
  6{x^3} + 14{x^2} \\
 }}}
\limits^{\displaystyle \,\,\, {2{x^2}}}\]
On subtracting this we get,
\[3x + 7\mathop{\left){\vphantom{1
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
 }}}
\limits^{\displaystyle \,\,\, {2{x^2}}}\]
Now let us multiply \[(3x + 7)\]by \[5x\]we get \[15{x^2} + 35x\]on subtracting this from the dividend we will easily eliminate the first term from that.
\[3x + 7\mathop{\left){\vphantom{1
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
  \underline {{\text{ }} - \left( {9x + 21} \right)} \\
  {\text{ }}0 \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
  \underline {{\text{ }} - \left( {9x + 21} \right)} \\
  {\text{ }}0 \\
 }}}
\limits^{\displaystyle \,\,\, {2{x^2} + 5x + 3}}\]
On subtracting this we get,
\[3x + 7\mathop{\left){\vphantom{1
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
 }}}
\limits^{\displaystyle \,\,\, {2{x^2} + 5x}}\]
Now let us multiply \[(3x + 7)\] by \[3\] we get \[9x + 21\] on subtracting this from the dividend we will easily eliminate the first term from that.
\[3x + 7\mathop{\left){\vphantom{1
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
  {\text{ }}9x + 21 \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
  {\text{ }}9x + 21 \\
 }}}
\limits^{\displaystyle \,\,\, {2{x^2} + 5x + 3}}\]
On subtracting this we get,
\[3x + 7\mathop{\left){\vphantom{1
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
  \underline {{\text{ }} - \left( {9x + 21} \right)} \\
  {\text{ }}0 \\
 }}\right.
\!\!\!\!\overline{\,\,\,\vphantom 1{
  {\text{ }}6{x^3} + 29{x^2} + 44x + 21 \\
  \underline { - \left( {6{x^3} + 14{x^2}} \right){\text{ }}} \\
  {\text{ }}15{x^2} + 44x \\
  \underline {{\text{ }} - \left( {15{x^2} + 35x} \right){\text{ }}} \\
  {\text{ }}9x + 21 \\
  \underline {{\text{ }} - \left( {9x + 21} \right)} \\
  {\text{ }}0 \\
 }}}
\limits^{\displaystyle \,\,\, {2{x^2} + 5x + 3}}\]
Now we got the remainder as zero. When the number or expression is divided by its factor it will give zero remainders. Thus, \[(3x + 7)\]is the factor of the given expression\[(6{x^3} + 29{x^2} + 44x + 21)\].
Hence proved.

Note:When a number or an expression (dividend) divided by another number or an expression (divisor) gives remainder zero then the number or an expression (divisor) which we used to divide is called the factor of the number or an expression (dividend) which is got divided.