Exploring Algebraic Identities | EOT

Question 1

Use suitable identities to find the following products:

(i) (−3x+4)2(-3x + 4)^2

(ii) (2s+7)(2s−7)(2s + 7)(2s - 7)

(iii) (p2+12)(p2−12)\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)

(iv) (2n+7)(2n−7)(2n + 7)(2n - 7)

(v) (s−2t)(s2+2st+4t2)(s - 2t)(s^2 + 2st + 4t^2)

(vi) (12r−4r)2\left(\dfrac{1}{2r} - 4r\right)^2

(vii) (−3m+4k−l)2(-3m + 4k - l)^2

(viii) (x−13y)3\left(x - \dfrac{1}{3}y\right)^3

(ix) (72k−23m)3\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3

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Solution
Understand the Question

To find the product of algebraic expressions efficiently, identify the matching standard algebraic identity and substitute the corresponding terms:

  • Square of a Binomial: (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 and (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2
  • Difference of Squares: (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2
  • Difference of Cubes: (a−b)(a2+ab+b2)=a3−b3(a - b)(a^2 + ab + b^2) = a^3 - b^3
  • Square of a Trinomial: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
  • Cube of a Binomial: (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

(i) Find the product: (−3x+4)2(-3x + 4)^2

Step 1 · Apply Binomial Square Identity

Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 with a=−3xa = -3x and b=4b = 4

(−3x+4)2=(−3x)2+2(−3x)(4)+42=9x2−24x+16\begin{aligned} (-3x + 4)^2 &= (-3x)^2 + 2(-3x)(4) + 4^2 \\[0.6em] &= 9x^2 - 24x + 16 \end{aligned}
Answer

(i) 9x2−24x+169x^2 - 24x + 16

(ii) Find the product: (2s+7)(2s−7)(2s + 7)(2s - 7)

Step 1 · Apply Difference of Squares Identity

Using the identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2 with a=2sa = 2s and b=7b = 7

(2s+7)(2s−7)=(2s)2−72=4s2−49\begin{aligned} (2s + 7)(2s - 7) &= (2s)^2 - 7^2 \\[0.6em] &= 4s^2 - 49 \end{aligned}
Answer

(ii) 4s2−494s^2 - 49

(iii) Find the product: (p2+12)(p2−12)\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)

Step 1 · Apply Difference of Squares Identity

Using the identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2 with a=p2a = p^2 and b=12b = \dfrac{1}{2}

(p2+12)(p2−12)=(p2)2−(12)2=p4−14\begin{aligned} \left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right) &= (p^2)^2 - \left(\frac{1}{2}\right)^2 \\[0.6em] &= p^4 - \frac{1}{4} \end{aligned}
Answer

(iii) p4−14p^4 - \dfrac{1}{4}

(iv) Find the product: (2n+7)(2n−7)(2n + 7)(2n - 7)

Step 1 · Apply Difference of Squares Identity

Using the identity (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2 with a=2na = 2n and b=7b = 7

(2n+7)(2n−7)=(2n)2−72=4n2−49\begin{aligned} (2n + 7)(2n - 7) &= (2n)^2 - 7^2 \\[0.6em] &= 4n^2 - 49 \end{aligned}
Answer

(iv) 4n2−494n^2 - 49

(v) Find the product: (s−2t)(s2+2st+4t2)(s - 2t)(s^2 + 2st + 4t^2)

Step 1 · Apply Difference of Cubes Identity

Rewrite the expression to match (a−b)(a2+ab+b2)=a3−b3(a - b)(a^2 + ab + b^2) = a^3 - b^3 where a=sa = s and b=2tb = 2t

(s−2t)(s2+2st+4t2)=(s−2t)[s2+s(2t)+(2t)2]=s3−(2t)3=s3−8t3\begin{aligned} (s - 2t)(s^2 + 2st + 4t^2) &= (s - 2t)\left[s^2 + s(2t) + (2t)^2\right] \\[0.6em] &= s^3 - (2t)^3 \\[0.6em] &= s^3 - 8t^3 \end{aligned}
Answer

(v) s3−8t3s^3 - 8t^3

(vi) Find the product: (12r−4r)2\left(\dfrac{1}{2r} - 4r\right)^2

Step 1 · Apply Binomial Square Identity

Using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 with a=12ra = \dfrac{1}{2r} and b=4rb = 4r

(12r−4r)2=(12r)2−2(12r)(4r)+(4r)2=14r2−4+16r2\begin{aligned} \left(\frac{1}{2r} - 4r\right)^2 &= \left(\frac{1}{2r}\right)^2 - 2\left(\frac{1}{2r}\right)(4r) + (4r)^2 \\[0.8em] &= \frac{1}{4r^2} - 4 + 16r^2 \end{aligned}
Answer

(vi) 14r2−4+16r2\dfrac{1}{4r^2} - 4 + 16r^2

(vii) Find the product: (−3m+4k−l)2(-3m + 4k - l)^2

Step 1 · Apply Trinomial Square Identity

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 with a=−3ma = -3m, b=4kb = 4k, and c=−lc = -l

(−3m+4k−l)2=[(−3m)+4k+(−l)]2=(−3m)2+(4k)2+(−l)2+2(−3m)(4k)+2(4k)(−l)+2(−l)(−3m)=9m2+16k2+l2−24mk−8kl+6lm\begin{aligned} (-3m + 4k - l)^2 &= [(-3m) + 4k + (-l)]^2 \\[0.6em] &= (-3m)^2 + (4k)^2 + (-l)^2 + 2(-3m)(4k) + 2(4k)(-l) + 2(-l)(-3m) \\[0.6em] &= 9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm \end{aligned}
Answer

(vii) 9m2+16k2+l2−24mk−8kl+6lm9m^2 + 16k^2 + l^2 - 24mk - 8kl + 6lm

(viii) Find the product: (x−13y)3\left(x - \dfrac{1}{3}y\right)^3

Step 1 · Apply Binomial Cube Identity

Using the identity (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 with a=xa = x and b=13yb = \dfrac{1}{3}y

(x−13y)3=x3−3(x)2(13y)+3(x)(13y)2−(13y)3=x3−3x2(13y)+3x(19y2)−127y3=x3−x2y+13xy2−127y3\begin{aligned} \left(x - \frac{1}{3}y\right)^3 &= x^3 - 3(x)^2\left(\frac{1}{3}y\right) + 3(x)\left(\frac{1}{3}y\right)^2 - \left(\frac{1}{3}y\right)^3 \\[0.8em] &= x^3 - 3x^2\left(\frac{1}{3}y\right) + 3x\left(\frac{1}{9}y^2\right) - \frac{1}{27}y^3 \\[0.8em] &= x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3 \end{aligned}
Answer

(viii) x3−x2y+13xy2−127y3x^3 - x^2y + \dfrac{1}{3}xy^2 - \dfrac{1}{27}y^3

(ix) Find the product: (72k−23m)3\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3

Step 1 · Apply Binomial Cube Identity

Using the identity (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 with a=72ka = \dfrac{7}{2}k and b=23mb = \dfrac{2}{3}m

(72k−23m)3=(72k)3−3(72k)2(23m)+3(72k)(23m)2−(23m)3=3438k3−3(494k2)(23m)+3(72k)(49m2)−827m3=3438k3−492k2m+143km2−827m3\begin{aligned} \left(\frac{7}{2}k - \frac{2}{3}m\right)^3 &= \left(\frac{7}{2}k\right)^3 - 3\left(\frac{7}{2}k\right)^2\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{2}{3}m\right)^2 - \left(\frac{2}{3}m\right)^3 \\[0.8em] &= \frac{343}{8}k^3 - 3\left(\frac{49}{4}k^2\right)\left(\frac{2}{3}m\right) + 3\left(\frac{7}{2}k\right)\left(\frac{4}{9}m^2\right) - \frac{8}{27}m^3 \\[0.8em] &= \frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3 \end{aligned}
Answer

(ix) 3438k3−492k2m+143km2−827m3\dfrac{343}{8}k^3 - \dfrac{49}{2}k^2m + \dfrac{14}{3}km^2 - \dfrac{8}{27}m^3

Common Mistakes
  • Sign Errors in Expansions: Forgetting that (−a)2=a2(-a)^2 = a^2 or mishandling signs when expanding trinomials like (−3m+4k−l)2(-3m + 4k - l)^2.
  • Power of Fractions and Coefficients: Failing to raise both the numerator, denominator, and variable to the power, e.g. writing (72k)3\left(\dfrac{7}{2}k\right)^3 as 3432k3\dfrac{343}{2}k^3 instead of 3438k3\dfrac{343}{8}k^3.
  • Middle Terms in Cubes: Miscalculating intermediate product terms like 3a2b3a^2b and 3ab23ab^2 due to incomplete fraction cancellations.

More questions in EOT

Q1

Use suitable identities to find the following products:

(i) (−3x+4)2(-3x + 4)^2

(ii) (2s+7)(2s−7)(2s + 7)(2s - 7)

(iii) (p2+12)(p2−12)\left(p^2 + \dfrac{1}{2}\right)\left(p^2 - \dfrac{1}{2}\right)

(iv) (2n+7)(2n−7)(2n + 7)(2n - 7)

(v) (s−2t)(s2+2st+4t2)(s - 2t)(s^2 + 2st + 4t^2)

(vi) (12r−4r)2\left(\dfrac{1}{2r} - 4r\right)^2

(vii) (−3m+4k−l)2(-3m + 4k - l)^2

(viii) (x−13y)3\left(x - \dfrac{1}{3}y\right)^3

(ix) (72k−23m)3\left(\dfrac{7}{2}k - \dfrac{2}{3}m\right)^3

Q2

Find the values using suitable identities:

(i) 17×2117 \times 21

(ii) 104×96104 \times 96

(iii) 24×1624 \times 16

(iv) 1473147^3

(v) 1993199^3

(vi) 1273127^3

(vii) (−107)3(-107)^3

(viii) (−299)3(-299)^3

Q3

Factor the following algebraic expressions:

(i) 4y2+1+116y24y^2 + 1 + \dfrac{1}{16y^2}

(ii) 9m2−125n29m^2 - \dfrac{1}{25n^2}

(iii) 27b3−164b327b^3 - \dfrac{1}{64b^3}

(iv) x2+5x6+16x^2 + \dfrac{5x}{6} + \dfrac{1}{6}

(v) 27u3−1125−27u25+9u2527u^3 - \dfrac{1}{125} - \dfrac{27u^2}{5} + \dfrac{9u}{25}

(vi) 64y3+1125z364y^3 + \dfrac{1}{125}z^3

(vii) p3+27q3+r3−9pqrp^3 + 27q^3 + r^3 - 9pqr

(viii) 9m2−12m+49m^2 - 12m + 4

(ix) 9x3−83y3+z33+6xyz9x^3 - \dfrac{8}{3}y^3 + \dfrac{z^3}{3} + 6xyz

(x) 4x2+9y2+36z2+12xz+36yz+24xy4x^2 + 9y^2 + 36z^2 + 12xz + 36yz + 24xy

(xi) 27u3−1216−9u22+u427u^3 - \dfrac{1}{216} - \dfrac{9u^2}{2} + \dfrac{u}{4}

Q4

Simplify the following:

(i) 4x2+4x+14x2−1\dfrac{4x^2 + 4x + 1}{4x^2 - 1}

(ii) 9(3a3−24b3)9a2−36b2\dfrac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}

(iii) s3+125t3s2−2st−35t2\dfrac{s^3 + 125t^3}{s^2 - 2st - 35t^2}

Note: Assume that the denominators are not equal to 0.

Q5

Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.

(i) 25a2−30ab+9b225a^2 - 30ab + 9b^2

(ii) 36s2−49t236s^2 - 49t^2

Q6

Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i) 6a2−24b26a^2 - 24b^2

(ii) 3ps2−15ps+12p3ps^2 - 15ps + 12p

Q7

The village playground is shaped as a square of side 40 metres. A path of width ss metres is created around the playground for people to walk. Find an expression for the area of the path in terms of ss.

Q8

If a number plus its reciprocal equals 103\dfrac{10}{3}, find the number.

Q9

A rectangular pool has area 2x2+7x+32x^2 + 7x + 3 square hastas. If its width is 2x+12x + 1 hastas, find its length. Hasta was a unit used to measure length.

Q10

If both x−2x - 2 and x−12x - \dfrac{1}{2} are factors of px2+5x+rpx^2 + 5x + r, show that p=rp = r.

Q11

If a+b+c=5a + b + c = 5 and ab+bc+ca=10ab + bc + ca = 10, then prove that a3+b3+c3−3abc=−25a^3 + b^3 + c^3 - 3abc = -25.

Q12

By factoring the expression, check that n3−nn^3 - n is always divisible by 6 for all natural numbers nn. Give reasons.

Q13

Find the value of

(i) x3+y3−12xy+64x^3 + y^3 - 12xy + 64, when x+y=−4x + y = -4

(ii) x3−8y3−36xy−216x^3 - 8y^3 - 36xy - 216, when x=2y+6x = 2y + 6

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