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Chapter 3: Problem 74

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

Short Answer

Expert verified
The value of \( k \) such that \( 4x + 3 \) is a factor of \( 20 x^{3}+23 x^{2}-10 x+k \) is \( k = \frac{693}{64} \).

Step by step solution

01

Setup Polynomial Division

Set up the division as would normally be done in polynomial division. Divide \( 20 x^{3}+23 x^{2}-10 x+k \) by \( 4x + 3 \) .
02

Substitute the Zero of the Factor

Since \( 4x + 3 \) is a possible factor, then \( x = - \frac{3}{4} \) is a possible zero of the cubic polynomial. Substitute this value into \( 20 x^{3}+23 x^{2}-10 x+k \) and set it equal to zero to solve for \( k \). So, \( 20(- \frac{3}{4})^3+23(- \frac{3}{4})^2-10(- \frac{3}{4})+k = 0 \).
03

Solve for k

The equation from Step 2 will simplify to \( k = \frac{693}{64} \).

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