Problem 62 - College Algebra

Question

When \(2 x^{2}-7 x+9\) is divided by a polynomial, the quotient is \(2 x-3\) and the remainder is \(3 .\) Find the polynomial.

Step-by-Step Solution

Answer

The polynomial we were looking for is \(x-2\).

Step 1: Understand the Dividend-Divisor-Quotient-Remainder Relations

It's important to know that the following relation holds for dividing polynomials: the dividend equals the product of divisor and quotient plus the remainder.

Step 2: Setup the Equation

Knowing that our dividend is \(2 x^{2}-7 x+9\), our quotient is \(2x-3\), and remainder is \(3\), we can set up the following equation: \(2 x^{2}-7 x+9 = d(x) \cdot (2x-3) + 3\), where \(d(x)\) is the polynomial we are trying to find.

Step 3: Solve the Equation

To solve this equation for \(d(x)\), rearrange it to find \(d(x)\): \(d(x) = \frac{2 x^{2}-7 x+9 - 3}{2x-3}\). Perform the subtraction in the numerator to get: \(d(x) = \frac{2 x^{2}-7 x+6}{2x-3}\)

Step 4: Simplify the Result

The expression in the numerator can be factorized, which gives us: \(d(x) = \frac{(2x-3) (x-2)}{2x-3}\). As \(2x-3\) appears in both numerator and denominator, they can be cancelled to get the polynomial \(d(x) = x-2\)