If a quadratic polynomial is of the form \(a^2x^2-b^2\), then

e.g.,

Can we factor the “difference of perfect cubes” using a similar rule?

Factor \(x^3-125\).

Solution

Let \(P(x)=x^3-125\). Then \(P(5)=(5)^3-125=0\). Therefore, \((x-5)\) is a factor.

Notice this factor is the difference in the cube roots of the terms of the expression we are factoring.

The quadratic factor can be obtained by long division, synthetic division, or the “have and need” method, which we will use here.

The first term of the quadratic factor must be \(x^2\) to obtain the \(x^3\) and the last term must be \(25\) to obtain the \(-125\).

Expanding what we have so far,

Have: \(x^3-5x^2+25x-125\)

Need: \(5x^2-25x\) \(=5x(x-5)\)

Hence, the middle term must be \(5x\).

This factoring is complete since the quadratic factor cannot be factored further.

Factor \(8x^3+27\).

Solution

Setting \(8x^3+27=0\) gives \(x=\sqrt[3]{-\frac{27}{8}}=-\frac{3}{2}\), so we could use \(\left(x+\frac{3}{2}\right)\) or \((2x+3)\) as the first factor.

Note that the terms of \((2x+3)\) are the cube roots of the terms of \(8x^3+27\).

The first term of the quadratic factor must be \(4x^2\) to obtain the \(8x^3\) and the last term must be \(9\) to obtain the \(27\).

Expand what we have so far.

Have: \(8x^3+12x^2+18x+27\)

Need: \(-12x^2-18x=-6x(2x+3)\)

Hence the middle term must be \(-6x\).

The factoring is complete since the quadratic factor cannot be factored further. This will always be the case when factoring sum and difference of cubes.

There are formulas that can be applied to factor sum and difference of cubes.

We can develop these by applying the steps used in the previous examples to a general form of the polynomial.

To develop the factoring formula for the difference of cubes, we will start with \(a^3x^3-b^3\).

It can be shown that this expression will equal zero when \(x=\sqrt[3]{\frac{b^3}{a^3}}\), therefore \(x=\frac{b}{a}\) so \((ax-b)\) can be used as the first factor.

The first term of the quadratic factor must be \(a^2x^2\) to obtain the \(a^3x^3\) and the last term must be \(b^2\) to obtain the \(b^3\).

Expanding what we have so far,

Have: \(a^3x^3-a^2bx^2+ab^2x-b^3\)

Need: \(a^2bx^2-ab^2x=abx(ax-b)\)

Hence, the middle term must be \(abx\).

Similarly, one can show that

Example 6

Factor \(64x^3-1=(4x)^3-(1)^3\).

Solution

Note that \(A=4x\) and \(B=1\), so

Factor the following polynomial:

$\mathrm{(((a)*(A))*3.0)*(d)}{\mathit{x}}^3\mathrm{(((p)*(m))*1.0)*(A)(((b)*(A))*3.0)*(s)}\mathrm{}$

 

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