# Video: Dividing Third-Degree Polynomials Using Long Division

Knowing that the volume of a box is 12π₯Β³ + 20π₯Β² β 21π₯ β 36, its length is 2π₯ + 3, and its width is 3π₯ β 4, express the height of the box algebraically.

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### Video Transcript

Knowing that the volume of a box is 12π₯ cubed plus 20π₯ squared minus 21π₯ minus 36, its length is two π₯ plus three, and its width is three π₯ minus four, express the height of the box algebraically.

We assume here that this box is in the shape of a cuboid. So its volume will be equal to its length multiplied by its width multiplied by its height. Weβve been given an algebraic expression for the volume, 12π₯ cubed plus 20π₯ squared minus 21 π₯ minus 36, as well as algebraic expressions for the length and width of the box, two π₯ plus three and three π₯ minus four. So we can substitute each of these expressions into the volume formula. To find the height of this box, we need to divide both sides of this equation by the expressions two π₯ plus three and three π₯ minus four, giving the quotient on the left of the screen.

Weβre going to need to simplify this quotient using polynomial division. But before we do, letβs expand the brackets in the denominator. Two π₯ multiplied by three π₯ gives six π₯ squared. Two π₯ multiplied by negative four gives negative eight π₯. Positive three multiplied by three π₯ gives positive nine π₯. And positive three multiplied by negative four gives negative 12. We can simplify by grouping the like terms in the center of the expansion to give six π₯ squared plus π₯ minus 12. So the height of the box is equal to the quotient 12π₯ cubed plus 20π₯ squared minus 21π₯ minus 36 over six π₯ squared plus π₯ minus 12. And we use algebraic division to simplify.

We set up our division with the divisor on the outside and the expression weβre dividing into on the inside. We begin by looking at the highest powers of these two expressions. And we ask ourselves, βWhat do we need to multiply six π₯ squared by to give 12 π₯ cubed?β Well, we need to multiply six by two to give 12 and π₯ squared by π₯ to give π₯ cubed. So overall, we must multiply by two π₯.

We now multiply the full expression six π₯ squared plus π₯ minus 12 by two π₯. Six π₯ squared multiplied by two π₯ gives 12π₯ cubed, as already discussed. π₯ multiplied by two π₯ gives two π₯ squared. And negative 12 multiplied by two π₯ gives negative 24π₯. Next, we subtract this expression from the original expression. 12π₯ cubed minus 12π₯ cubed gives zero. 20π₯ squared minus two π₯ squared gives 18π₯ squared. And negative 21π₯ minus negative 24π₯ becomes negative 21π₯ plus 24π₯, which is positive three π₯. We also have negative 36 minus zero. So, we still have negative 36.

We still have a remainder at this point. So, we keep going. And again, we look at the highest powers, asking ourselves, βWhat do we multiply six π₯ squared by to give 18π₯ squared?β We must multiply by three. Six multiplied by three is 18. So six π₯ squared multiplied by three is 18π₯ squared. Now we multiply the full expression by three. Six π₯ squared multiplied by three gives 18π₯ squared. π₯ multiplied by three gives plus three π₯. And negative 12 multiplied by three gives negative 36.

As before, we subtract in columns. 18π₯ squared minus 18π₯ squared gives zero. Three π₯ minus three π₯ gives zero. And negative 36 minus negative 36 β thatβs negative 36 plus 36 β is also zero. So, we have no remainder. And weβve therefore reached the end of our algebraic division. In total, we multiplied six π₯ squared plus π₯ minus 12 by two π₯ and then by positive three. So overall, weβve multiplied by two π₯ plus three. This tells us that the answer to the algebraic division 12π₯ cubed plus 20π₯ squared minus 21π₯ minus 36 divided by six π₯ squared plus π₯ minus 12 is two π₯ plus three. And so this is the height of this box expressed algebraically. The height is two π₯ plus three.