Exploring Algebraic Identities | Exercise 4.3

Question 3

Expand the following using the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca:

(i) (p+3q+7r)2(p + 3q + 7r)^2

(ii) (3x−2y+4z)2(3x - 2y + 4z)^2

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Solution
Understand the Question
  • We expand algebraic trinomial squares using the standard algebraic identity: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
  • To apply this formula:
    1. Compare the given expression to (a+b+c)2(a + b + c)^2 to identify the values of aa, bb, and cc.
    2. For terms with negative signs, rewrite them in additive form, such as [3x+(−2y)+4z]2[3x + (-2y) + 4z]^2, so that b=−2yb = -2y.
    3. Substitute these values into the identity and simplify.

(i) Expand (p+3q+7r)2(p + 3q + 7r)^2

Step 1 · Apply Identity and Expand

Diagram 2

Using the identity: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Here, a=pa = p, b=3qb = 3q, and c=7rc = 7r.

(p+3q+7r)2=(p)2+(3q)2+(7r)2+2(p)(3q)+2(3q)(7r)+2(p)(7r)=p2+9q2+49r2+6pq+42qr+14pr\begin{aligned} (p + 3q + 7r)^2 &= (p)^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(p)(7r) \\[0.6em] &= p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr \end{aligned}
Answer

(i) p2+9q2+49r2+6pq+42qr+14prp^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr

(ii) Expand (3x−2y+4z)2(3x - 2y + 4z)^2

Step 1 · Apply Identity and Expand

Rewrite the expression in additive form: (3x−2y+4z)2=[3x+(−2y)+4z]2(3x - 2y + 4z)^2 = [3x + (-2y) + 4z]^2

Using the identity: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Here, a=3xa = 3x, b=−2yb = -2y, and c=4zc = 4z.

(3x−2y+4z)2=(3x)2+(−2y)2+(4z)2+2(3x)(−2y)+2(−2y)(4z)+2(3x)(4z)=9x2+4y2+16z2−12xy−16yz+24xz\begin{aligned} (3x - 2y + 4z)^2 &= (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(3x)(4z) \\[0.6em] &= 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz \end{aligned}
Answer

(ii) 9x2+4y2+16z2−12xy−16yz+24xz9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz

Common Mistakes
  • Squaring Negative Terms: Squaring any term always gives a positive value, so (−2y)2=+4y2(-2y)^2 = +4y^2, not −4y2-4y^2.
  • Sign Errors in Cross-Products: When multiplying positive and negative terms, remember the signs: 2(3x)(−2y)=−12xy2(3x)(-2y) = -12xy and 2(−2y)(4z)=−16yz2(-2y)(4z) = -16yz.
  • Forgetting the Coefficient 2: Don't forget to multiply the mixed product terms by 22 (writing abab instead of 2ab2ab).

More questions in Exercise 4.3

Q1

Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.

(i) 1172117^2

(ii) 78278^2

(iii) 1982198^2

(iv) 2142214^2

(v) 110421104^2

(vi) 112021120^2

Q2

Factor using suitable identities:

(i) 16y2−24y+916y^2 - 24y + 9

(ii) 94s2+6st+4t2\dfrac{9}{4}s^2 + 6st + 4t^2

(iii) m29+mk3+k24+3nk+2mn+9n2\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2

(iv) p216−2+16p2\dfrac{p^2}{16} - 2 + \dfrac{16}{p^2}

(v) 9a2+4b2+c2−12ab+6ac−4bc9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc

Q3

Expand the following using the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca:

(i) (p+3q+7r)2(p + 3q + 7r)^2

(ii) (3x−2y+4z)2(3x - 2y + 4z)^2

Q4

Is this an identity?

(a+b−c)2+(a−b+c)2+(a−b−c)2=2a2+2b2+2c2(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2

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