Factor the polynomial with integer coefficients given that \( f(k)=0 \) a. \( f(x)=3 x^{3}-16 x^{2}+3 x+10 ; f(5)=0(x-1)(x-1 \) b. \( f(x)=3 x^{4}+5 x^{3}+20 x-48 ; f(2 i)=0\left(x^{2}+2\right) \)

To factor the given polynomials, we start with the polynomial in part a. For \( f(x) = 3x^3 - 16x^2 + 3x + 10 \) and knowing \( f(5) = 0 \), we can use synthetic division to factor \( f(x) \) by \( (x - 5) \). Upon performing synthetic division, we find that: \[ f(x) = (x - 5)(3x^2 - 1) \] Next, we can factor \( 3x^2 - 1 \) further using the difference of squares: \[ 3x^2 - 1 = \sqrt{3} x - 1)(\sqrt{3} x + 1) \] Thus, the complete factorization of part a is: \[ f(x) = (x - 5)(\sqrt{3} x - 1)(\sqrt{3} x + 1) \] For part b where \( f(x) = 3x^4 + 5x^3 + 20x - 48 \) and knowing \( f(2i) = 0 \), we first note that complex roots come in conjugate pairs, so \( f(-2i) = 0 \) as well. We can factor out \( (x^2 + 4) \) from \( f(x) \) using synthetic division leading to the quotient: \[ f(x) = (x^2 + 4)(3x^2 + 5x - 12) \] Next, we can factor \( 3x^2 + 5x - 12 \) by finding two numbers that multiply to \( -36 \) (the product of \( 3 \) and \( -12 \)) and add to \( 5 \). The pair \( 9 \) and \( -4 \) works, and thus we have: \[ 3x^2 + 5x - 12 = (3x - 4)(x + 3) \] So, the complete factorization of part b is: \[ f(x) = (x^2 + 4)(3x - 4)(x + 3) \] To summarize: a. \( f(x) = (x - 5)(\sqrt{3} x - 1)(\sqrt{3} x + 1) \) b. \( f(x) = (x^2 + 4)(3x - 4)(x + 3) \)