6.01c–6.01e - More demanding algebraic products

6.01c–6.01e - More demanding algebraic products

Fractional powers, extra brackets and a coefficient before a squared term all make algebra look less familiar. You will extend index laws in 6.01c, expand several binomials in 6.01d and factorise general quadratics in 6.01e. The same habit helps throughout: recognise the structure before doing the arithmetic.

Algebraic terms with fractional powers

Can a product involving square roots simplify to a term with no root at all?

Higher only. Try a numerical case: 91/2×91/2=3×3=99^{1/2}\times9^{1/2}=3\times3=9. The indices add to 11, just as they do for whole-number powers. Letters follow the same laws when the powers are defined.

Fractional and negative powers

For a positive base xx, x1/2=xx^{1/2}=\sqrt{x}, x3/2=(x)3x^{3/2}=(\sqrt{x})^3 and x−1/2=1/xx^{-1/2}=1/\sqrt{x}. A negative index takes a reciprocal; it does not make the value negative.

For positive bases, you may add indices when multiplying the same base, subtract them when dividing, and multiply them when taking a power of a power:

xpxq=xp+q,xpxq=xp−q,(xp)q=xpq.x^p x^q=x^{p+q},\qquad \frac{x^p}{x^q}=x^{p-q},\qquad (x^p)^q=x^{pq}.

Here pp and qq can be fractions or negative numbers. Deal with numerical coefficients separately. Our examples use positive variables so that all the fractional powers are unambiguous real numbers. Zero is allowed in many positive powers, but not in a denominator or a negative power. Some odd-root expressions also allow negative bases; that does not justify using every fractional-power rule indiscriminately on negative bases.

Worked example 1: adding fractional indices

Question. Simplify 6x3/2×2x−1/26x^{3/2}\times2x^{-1/2}, where x>0x>0.

Plan. Multiply the coefficients and add the indices of xx.

6x3/2×2x−1/2=12x3/2+(−1/2)same base=12x1add the fractions=12xsimplify\begin{aligned} 6x^{3/2}\times2x^{-1/2} &=12x^{3/2+(-1/2)} && \text{same base} \\ &=12x^1 && \text{add the fractions} \\ &=12x && \text{simplify} \end{aligned}

Answer. 12x12x.

Check. Write the powers as roots: 6xx×2/x=12x6x\sqrt{x}\times2/\sqrt{x}=12x. The root cancels because x>0x>0.

Change only multiplication to division. Now the second index is subtracted, including its negative sign.

Worked example 2: subtracting a negative index

Question. Simplify 6x3/2÷2x−1/26x^{3/2}\div2x^{-1/2}, where x>0x>0.

Plan. Divide the coefficients and subtract the denominator's index.

6x3/22x−1/2=3x3/2−(−1/2)divide like bases=3x2subtract the negative\begin{aligned} \frac{6x^{3/2}}{2x^{-1/2}} &=3x^{3/2-(-1/2)} && \text{divide like bases} \\ &=3x^2 && \text{subtract the negative} \end{aligned}

Answer. 3x23x^2.

Check. Dividing 6xx6x\sqrt{x} by 2/x2/\sqrt{x} gives 3xxx=3x23x\sqrt{x}\sqrt{x}=3x^2.

Power-of-a-power questions use a different operation: for example, (4x1/2)2=16x(4x^{1/2})^2=16x, because the outside square applies to both factors and (1/2)×2=1(1/2)\times2=1.

Watch out

The line x3/2÷x−1/2=xx^{3/2}\div x^{-1/2}=x adds the indices instead of subtracting them. Write 3/2−(−1/2)=23/2-(-1/2)=2 before simplifying.

Keep the coefficient calculation separate in your own answer.

Expanding three brackets

Three brackets are still a product, so you can multiply two of them first and then multiply the result by the third.

There is no new expansion rule. Every term in one factor must multiply every term in the other. Choose a convenient pair first: (x+2)(x−2)=x2−4(x+2)(x-2)=x^2-4 is shorter than a general quadratic because its middle terms cancel.

Binomial

A binomial has two terms, such as x+2x+2 or 3x−13x-1. The product of three linear binomials normally has a highest power of x3x^3; collect like powers after expanding.

Keep the intermediate quadratic in brackets. Otherwise it is easy to multiply only its first term by the final factor.

Worked example 3: choosing the easiest pair

Question. Expand and simplify (x+2)(x−2)(x+3)(x+2)(x-2)(x+3).

Plan. Use the difference of two squares, then distribute the final bracket.

(x+2)(x−2)(x+3)=(x2−4)(x+3)first two factors=x3+3x2−4x−12multiply all four pairs\begin{aligned} (x+2)(x-2)(x+3) &=(x^2-4)(x+3) && \text{first two factors} \\ &=x^3+3x^2-4x-12 && \text{multiply all four pairs} \end{aligned}

Answer. x3+3x2−4x−12x^3+3x^2-4x-12.

Check. At x=2x=2, the product is zero. The expansion gives 8+12−8−12=08+12-8-12=0. Its leading term is also x×x×x=x3x\times x\times x=x^3, and its constant term is 2×(−2)×3=−122\times(-2)\times3=-12.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account