Use a vertical format to find the product of 4x^3 + x^2 + 2x + 6 and x + 5, then simplify.

Step-by-step solution

1. 1

   Understand what "vertical format" buys you. Laying the multiplication out like long multiplication of numbers means you generate two partial products — one from each term of the binomial — and then add them with like powers stacked in the same column. That column alignment is what prevents the classic error of combining an $x^{3}$ term with an $x^{2}$ term.

2. 2

   Form the first partial product by distributing $x$. Multiplying each term of the cubic by $x$ raises every exponent by one:

   $x\left(4x^{3}+x^{2}+2x+6\right)=4x^{4}+x^{3}+2x^{2}+6x$

3. 3

   Form the second partial product by distributing $5$. Here only the coefficients change:

   $5\left(4x^{3}+x^{2}+2x+6\right)=20x^{3}+5x^{2}+10x+30$

4. 4

   Add the two rows column by column. Group by power of $x$:

   $4x^{4}+\left(x^{3}+20x^{3}\right)+\left(2x^{2}+5x^{2}\right)+\left(6x+10x\right)+30$

   Note that $4x^{4}$ has no partner (only the $x$ row produced a quartic term) and $30$ has no partner (only the $5$ row produced a constant).

5. 5

   Combine and state the result.

   $4x^{4}+21x^{3}+7x^{2}+16x+30$

   A quick degree check: a degree-$3$ factor times a degree-$1$ factor must give degree $4$, and the leading coefficient must be $4\cdot 1=4$. Both hold.

6. 6

   Verify with test values. At $x=0$ both the factored form and the expanded form give $6\cdot 5=30$. At $x=1$ both give $13\cdot 6=78$, at $x=2.5$ both give $598.125$, and at $x=-1.3$ both give $-13.6826$. Agreement at five points pins down a degree-$4$ polynomial completely.