Problem 61 - College Algebra

Question

Find \(k\) so that \(4 x+3\) is a factor of $$20 x^{3}+23 x^{2}-10 x+k$$

Step-by-Step Solution

Answer

The value of \(k\) that makes \(4x+3\) a factor of \(20 x^{3}+23x^{2}-10x+k\) is -13.25. This value is obtained by equating the polynomial to zero for the value of \(x\) that nullifies the factor, and then solving for \(k\).

Step 1: Solve \(4x+3\) for \(x\)

First, solve the equation \(4x+3=0\) for \(x\) to obtain \(x=-3/4\). Here's the calculation: \(4x+3=0\) Subtraction 3 from both sides gives: \(4x=-3\) Dividing both sides by 4 gives: \(x=-3/4\)

Step 2: Plug \(x=-3/4\) into the polynomial

Next, you need to plug \(x=-3/4\) into the polynomial \(20 x^{3}+23x^{2}-10x+k\) and equate it to zero, since a factor should nullify the polynomial. \(20(-3/4)^3+23(-3/4)^2-10(-3/4)+k=0\)

Step 3: Compute and solve for \(k\)

Simplify the equation to find the value of \(k\) After mathematical manipulation, the equation simplifies to: \(-27+131/4+15/2+k=0\) Simplify further to: \(-27+32.75+7.5+k=0\) Which simplifies to: \(k= -13.25\) \ k is equal to negative 13.25