Question

the expression $x + 5$ is one factor of the polynomial $2x^3 + 17x^2 + 38x + 15$. use long division to find the remaining factors of the polynomial.
a $x - 3, 2x + 1$
b $x + 3, 2x + 1$
c $x - 3, 2x - 1$
d $x + 3, 2x - 1$

Explanation:

Step1: Set up long division

Divide \(2x^3 + 17x^2 + 38x + 15\) by \(x + 5\).
First term: \(\frac{2x^3}{x}=2x^2\). Multiply \(x + 5\) by \(2x^2\): \(2x^2(x + 5)=2x^3 + 10x^2\).
Subtract from the polynomial: \((2x^3 + 17x^2 + 38x + 15)-(2x^3 + 10x^2)=7x^2 + 38x + 15\).

Step2: Next term in division

Divide \(7x^2\) by \(x\) to get \(7x\). Multiply \(x + 5\) by \(7x\): \(7x(x + 5)=7x^2 + 35x\).
Subtract: \((7x^2 + 38x + 15)-(7x^2 + 35x)=3x + 15\).

Step3: Final term in division

Divide \(3x\) by \(x\) to get \(3\). Multiply \(x + 5\) by \(3\): \(3(x + 5)=3x + 15\).
Subtract: \((3x + 15)-(3x + 15)=0\).
So, \(2x^3 + 17x^2 + 38x + 15=(x + 5)(2x^2 + 7x + 3)\).

Step4: Factor the quadratic

Factor \(2x^2 + 7x + 3\). We need two numbers that multiply to \(2\times3 = 6\) and add to \(7\). The numbers are \(6\) and \(1\).
Rewrite: \(2x^2 + 6x + x + 3 = 2x(x + 3)+1(x + 3)=(x + 3)(2x + 1)\).
So the factors are \(x + 5\), \(x + 3\), and \(2x + 1\). The remaining factors (other than \(x + 5\)) are \(x + 3\) and \(2x + 1\).

Answer:

B. \(x + 3\), \(2x + 1\)