which polynomial has (3x + 2) as a binomial factor?\n6x³ + 3x² + 4x + 2\n12x² + 15x + 8x + 10\n18x³ - 12x² +…

Explanation:

Step1: Use the factor - theorem

If (3x + 2) is a factor, then (x=-\frac{2}{3}) is a root of the polynomial. We substitute (x =-\frac{2}{3}) into each polynomial.

Step2: Check the first polynomial (P_1(x)=6x^{3}+3x^{2}+4x + 2)

[ \begin{align*} P_1\left(-\frac{2}{3}\right)&=6\left(-\frac{2}{3}\right)^{3}+3\left(-\frac{2}{3}\right)^{2}+4\left(-\frac{2}{3}\right)+2\ &=6\times\left(-\frac{8}{27}\right)+3\times\frac{4}{9}-\frac{8}{3}+2\ &=-\frac{16}{9}+\frac{4}{3}-\frac{8}{3}+2\ &=-\frac{16}{9}+\frac{12 - 24+18}{9}\ &=-\frac{16}{9}+\frac{6}{9}\ &=-\frac{10}{9}\neq0 \end{align*} ]

Step3: Check the second polynomial (P_2(x)=12x^{2}+15x + 8x+10=12x^{2}+23x + 10)

[ \begin{align*} P_2\left(-\frac{2}{3}\right)&=12\left(-\frac{2}{3}\right)^{2}+23\left(-\frac{2}{3}\right)+10\ &=12\times\frac{4}{9}-\frac{46}{3}+10\ &=\frac{16}{3}-\frac{46}{3}+10\ &=\frac{16 - 46 + 30}{3}\ &=0 \end{align*} ] We can stop here since we have found the polynomial for which (x =-\frac{2}{3}) is a root. But for completeness, we can check the others.

Step4: Check the third polynomial (P_3(x)=18x^{3}-12x^{2}+9x - 6)

[ \begin{align*} P_3\left(-\frac{2}{3}\right)&=18\left(-\frac{2}{3}\right)^{3}-12\left(-\frac{2}{3}\right)^{2}+9\left(-\frac{2}{3}\right)-6\ &=18\times\left(-\frac{8}{27}\right)-12\times\frac{4}{9}-6 - 6\ &=-\frac{16}{3}-\frac{16}{3}-12\ &=\frac{-16-16 - 36}{3}\ &=-\frac{68}{3}\neq0 \end{align*} ]

Step5: Check the fourth polynomial (P_4(x)=21x^{4}+7x^{3}+6x + 2)

[ \begin{align*} P_4\left(-\frac{2}{3}\right)&=21\left(-\frac{2}{3}\right)^{4}+7\left(-\frac{2}{3}\right)^{3}+6\left(-\frac{2}{3}\right)+2\ &=21\times\frac{16}{81}+7\times\left(-\frac{8}{27}\right)-4 + 2\ &=\frac{112}{27}-\frac{56}{27}-2\ &=\frac{112 - 56-54}{27}\ &=\frac{2}{27}\neq0 \end{align*} ]

Answer:

(12x^{2}+15x + 8x + 10)