What is the other factor of the polynomial 2x² + 5x – 3 if one factor is x – 1?

To find the other factor of the polynomial 2x² + 5x – 3 given that one of the factors is x – 1, we can use polynomial long division or synthetic division.

First, we can rewrite 2x² + 5x – 3 as:

• (x – 1)(Ax + B)

We need to find the constants A and B. We can set up the equation:

2x² + 5x – 3 = (x – 1)(Ax + B)

Expanding the right side:

(x – 1)(Ax + B) = Ax² + Bx – Ax – B = Ax² + (B – A)x – B

Now we can equate the coefficients from both sides:

• A = 2 (coefficient of x²)
• B – A = 5 (coefficient of x)
• -B = -3 (constant term)

From -B = -3, we find B = 3. Substituting B into B – A = 5 gives us:

3 – 2 = 1 (which is not 5, so it seems I made an incorrect assumption in my explanation).

Let’s explicitly divide:

Step 1: Divide leading term: 2x² / x = 2x
Step 2: Multiply: (x - 1)(2x) = 2x² - 2x
Step 3: Subtract: (2x² + 5x - 3) - (2x² - 2x) = 7x - 3
Step 4: Divide leading term: 7x / x = 7
Step 5: Multiply: (x - 1)(7) = 7x - 7
Step 6: Subtract: (7x - 3) - (7x - 7) = 4

This means the quotient is 2x + 7 and the remainder is 4.

Thus, the correct way to express the factoring would be:

(x – 1)(2x + 7) + 4

This concludes that if we know that x – 1 is a factor, we cannot say it divides evenly since there’s a remainder. We can, therefore, conclude the polynomial does not fully factor to yield integer coefficients without keeping the remainder in mind.
So, if one wanted to express the polynomial:

The factored form relating closely would be 2x + 7 (but remainder needs to be added back)