Questions: Factor f(x) into linear factors given that k is a zero of f(x). f(x)=4x^3+7x^2-142x+35; k=5 f(x)=(x-5)(x-4)(x+7) (Factor completely.)

Solution Steps

To factor the polynomial \( f(x) = 4x^3 + 7x^2 - 142x + 35 \) given that \( k = 5 \) is a zero, we can use synthetic division to divide the polynomial by \( x - 5 \). This will give us a quadratic polynomial, which we can then factor further to find the remaining linear factors.

We start with the polynomial \( f(x) = 4x^3 + 7x^2 - 142x + 35 \) and perform synthetic division using the zero \( k = 5 \). The coefficients of the polynomial are \( [4, 7, -142, 35] \). After performing synthetic division, we obtain the new coefficients \( [4, 27, -7, 0] \), which represent the polynomial \( 4x^2 + 27x - 7 \).

Next, we need to find the roots of the quadratic polynomial \( 4x^2 + 27x - 7 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 27 \), and \( c = -7 \), we calculate the roots:

\[ x = \frac{-27 \pm \sqrt{27^2 - 4 \cdot 4 \cdot (-7)}}{2 \cdot 4} \]

Calculating the discriminant:

\[ 27^2 - 4 \cdot 4 \cdot (-7) = 729 + 112 = 841 \]

Thus, the roots are:

\[ x = \frac{-27 \pm \sqrt{841}}{8} = \frac{-27 \pm 29}{8} \]

Calculating the two roots:

1. \( x_1 = \frac{2}{8} = 0.25 \)
2. \( x_2 = \frac{-56}{8} = -7 \)

Now that we have the roots \( x = 5 \), \( x = 0.25 \), and \( x = -7 \), we can express the polynomial \( f(x) \) in its factored form:

\[ f(x) = (x - 5)(x - 0.25)(x + 7) \]

Final Answer