Exploring Algebraic Identities: questions with answers

The outer terms are $(4x)^{2}$ and $6^{2}$. The squares $(4x + 6)^{2}$ and $(4x - 6)^{2}$ give middle coefficients $48$ and $-48$, respectively. Since $k$ must be positive, use $(4x + 6)^{2}$. Thus $k = 2 \times 4 \times 6 = 48$.

Take $a = 2x$ and $b = 5y$ in $(a + b)^{2} = a^{2} + 2ab + b^{2}$. Squaring a term squares everything in it, so $a^{2}$ is $4x^{2}$ and $b^{2}$ is $25y^{2}$ — not $2x^{2}$ and $5y^{2}$. The cross term is $2(2x)(5y) = 20xy$, and it is there twice, once from each bracket, which is why the expansion has three terms and not two.

At $(a,b)=(7,1)$, the left side is $(7+1)^2=64$. The right side is $7^2+1^2=50$. These values differ, so the equation is not an identity. Agreement at one pair does not prove an identity.

Use the distributive property: $(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2$. This proves equality for all values of $a$ and $b$. Checking one pair, or a few pairs, proves agreement only at those values.

Expanding gives $(x + 2)(x + 10) = x^{2} + 12x + 20$. The constant term is already $20$, since $2 \times 10 = 20$. Only the middle term differs: the given expression has $9x$ and the product has $12x$. So the term to add is $12x - 9x = 3x$. Adding $0$ would mean the two are already equal. They are not: $2 + 10 = 12$, not $9$. A pair with the right product must also add to the middle coefficient.

The product expands to $x^{2} + 9x + 18$. The coefficient of $x$ matches, but the constant term differs. The constant to add is $20 - 18 = 2$. Matching the sum alone does not give the required factorisation.

The terms are squares: $49s^{2} = (7s)^{2}$ and $4t^{2} = (2t)^{2}$. Use $a^{2} - b^{2} = (a+b)(a-b)$. Here, $a = 7s$ and $b = 2t$. The factorisation is $(7s + 2t)(7s - 2t)$. Repeating either bracket introduces a middle term, $-28st$ or $+28st$. The original expression has no $st$ term.

The constant has to be recognised as a square first: $49 = 7^{2}$, and $16x^{2} = (4x)^{2}$. Then $a^{2} - b^{2} = (a+b)(a-b)$ gives $(4x + 7)(4x - 7)$. Note that it is the SQUARE ROOTS that go into the brackets, $4$ and $7$ — putting $16$ and $49$ there instead would multiply out to something far larger.

Use $(a-b)^{2}=a^{2}-2ab+b^{2}$ with $a=6x$ and $b=3y$. This gives $36x^{2}-36xy+9y^{2}$. The last term has a plus sign because $(-3y)^{2}=9y^{2}$. Omitting the middle term would leave $36x^{2}+9y^{2}$, which is incorrect. The difference $36x^{2}-9y^{2}$ instead equals $(6x+3y)(6x-3y)$.

Use $(a-b)^2=a^2-2ab+b^2$ with $a=5x$ and $b=9$. The middle term is $-2(5x)(9)=-90x$. The last term is $(-9)^2=+81$. Thus, the expansion is $25x^{2}-90x+81$. Dropping only the middle term would give $25x^{2}+81$. The expression $25x^{2}-81$ instead comes from confusing $(a-b)^2$ with $a^2-b^2$.

Square each term, then add twice the product of each pair of terms. The squared terms are $4x^{2}$, $16y^{2}$ and $9z^{2}$. The cross terms containing $-3z$ are $-24yz$ and $-12zx$. The remaining cross term is $+16xy$. Include all three pairs, including the first and last terms.

Every pair of the three parts contributes, and each contributes twice — three squares and three doubled products, six terms in all. Each cross term takes the product of its OWN two signs: $x$ with $y$ gives $-16xy$ and $y$ with $z$ gives $-40yz$, but $z$ with $x$ pairs two positives and gives $+20zx$. The $zx$ term is easy to lose because $z$ and $x$ are not written next to each other, yet in the 3-by-3 grid of Fig. 4.4 its two rectangles are as present as any other pair's.

$(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$, and the two 3s are what the cube picture counts: besides the corner cubes there are THREE cuboids of volume $a^{2}b$ and three of volume $ab^{2}$. So with $a = 2x$ and $b = 5y$ the middle coefficients are $3 \times 4 \times 5 = 60$ and $3 \times 2 \times 25 = 150$. The coefficient is 3, not the 2 that belongs to the square, and the middle terms do not disappear.

With $a = x$ and $b = 5$: $a^{3} = x^{3}$, $3a^{2}b = 15x^{2}$, $3ab^{2} = 75x$ and $b^{3} = 125$. Both middle coefficients carry a factor of 3 — one for each of the three cuboids of that shape — so they are $15$ and $75$, not $10$ and $50$. And a cube of a sum is never just the sum of the cubes: at $x = 1$ this expansion is $216$ while $1 + 125$ is $126$.

Factor before cancelling. The numerator is $(x + 3)^{2}$. The denominator is $(x + 3)(x + 8)$. Cancel one common factor $(x + 3)$ to get $\frac{x + 3}{x + 8}$. One copy remains in the numerator because it originally had two copies. You cannot cancel $x^{2}$ directly from the original sums: it is a term, not a common factor.