2. what is the quotient of $x^{3}+8x^{2}+x - 42$ and $x - 2$? use the quotient to find the remaining linear…

Explanation:

Step1: Use polynomial long - division

We divide $x^{3}+8x^{2}+x - 42$ by $x - 2$. First, divide the leading term of the dividend $x^{3}+8x^{2}+x - 42$ (which is $x^{3}$) by the leading term of the divisor $x - 2$ (which is $x$). The result is $x^{2}$. Multiply $x - 2$ by $x^{2}$ to get $x^{3}-2x^{2}$. Subtract this from the dividend: $(x^{3}+8x^{2}+x - 42)-(x^{3}-2x^{2})=10x^{2}+x - 42$.

Step2: Continue the long - division

Divide the leading term of $10x^{2}+x - 42$ (which is $10x^{2}$) by the leading term of $x - 2$ (which is $x$). The result is $10x$. Multiply $x - 2$ by $10x$ to get $10x^{2}-20x$. Subtract this from $10x^{2}+x - 42$: $(10x^{2}+x - 42)-(10x^{2}-20x)=21x - 42$.

Step3: Finish the long - division

Divide the leading term of $21x - 42$ (which is $21x$) by the leading term of $x - 2$ (which is $x$). The result is $21$. Multiply $x - 2$ by $21$ to get $21x-42$. Subtract this from $21x - 42$: $(21x - 42)-(21x - 42)=0$. So the quotient is $x^{2}+10x + 21$.

Step4: Factor the quotient

We factor $x^{2}+10x + 21$. We need to find two numbers that multiply to $21$ and add up to $10$. The numbers are $3$ and $7$. So $x^{2}+10x + 21=(x + 3)(x+7)$.

Answer:

The quotient is $x^{2}+10x + 21$ and the remaining linear factors are $x + 3$ and $x + 7$.