Question

Find the polynomial which when divided by \[3x + 4\], equals \[2{x^2} + 5x - 3\] with a remainder of \[3\]
a) \[6{x^3} + 23{x^2} - 11x - 12\]
b) \[6{x^3} + 23{x^2} - 11x - 9\]
c) \[6{x^3} + 23{x^2} - 11x - 15\]
d) \[6{x^3} + 23{x^2} + 11x - 12\]
e) \[6{x^3} + 23{x^2} + 11x - 9\]

Answer 1

Hint: In the given question we have to find the polynomial that is the dividend and we have the divisor, the quotient and the remainder given in the question. So, to find the polynomial that is the dividend we will use the formula that;
Dividend \[ = \] divisor \[ \times \] quotient \[ + \] remainder

Complete answer:
We have;
Divisor \[ = 3x + 4\]
Quotient \[ = 2{x^2} + 5x - 3\]
Remainder \[ = 3\]
Now to find the dividend we will use the formula;
Dividend \[ = \] divisor \[ \times \] quotient \[ + \] remainder
So, putting the value we have;
\[ \Rightarrow {\text{ dividend = }}\left( {3x + 4} \right)\left( {2{x^2} + 5x - 3} \right) + 3\]
Now we will expand the bracket. So, we have;
\[ \Rightarrow {\text{ dividend = 6}}{{\text{x}}^3} + 15{x^2} - 9x + 8{x^2} + 20x - 12 + 3\]
On further simplification we get;
\[ \Rightarrow {\text{ dividend = 6}}{{\text{x}}^3} + 23{x^2} + 11x - 9\]

Hence, the last option e is correct.

Additional information:
The degree of a polynomial is the highest exponent of the variable in the expression. Polynomials with degree as one are called linear polynomials. Polynomials with degree as two are called quadratic polynomials. The polynomials with degree as zero are called constant polynomials. We must remember that the degree of the remainder must be less than that of the divisor.

Note:
The property of long division: Dividend \[ = \] divisor \[ \times \] quotient \[ + \] remainder, holds true for division of numbers as well as polynomials. The degree of quotient given to us is two and the degree of the divisor is one. So, the degree of dividend must be three as the multiplication of a quadratic expression with a linear expression yields a cubic expression.