Exploring Algebraic Identities | Exercise 4.4

Question 3

Factor the following:

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

(ii) 16s2+25t2−40st16s^2 + 25t^2 - 40st

(iii) r2−r−42r^2 - r - 42

(iv) 49g2+14gh+h249g^2 + 14gh + h^2

(v) 64u2+121v2+4w2−176uv−32uw+44vw64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question

To factorize each algebraic expression, we use appropriate algebraic identities and factorization techniques:

  • Square of a Binomial: (a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2
  • Square of a Trinomial: (x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx
  • Splitting the Middle Term: For a quadratic trinomial x2+bx+cx^2 + bx + c, find two numbers whose product is cc and whose sum is bb.

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

Step 1 · Apply Identity for Square of a Trinomial

Compare with (x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx:

  • x2=9a2  ⟹  x=3ax^2 = 9a^2 \implies x = 3a
  • y2=b2y^2 = b^2 and 2xy=−6ab  ⟹  y=−b2xy = -6ab \implies y = -b
  • z2=4c2z^2 = 4c^2 and 2xz=12ac  ⟹  z=2c2xz = 12ac \implies z = 2c
  • Check: 2yz=2(−b)(2c)=−4bc2yz = 2(-b)(2c) = -4bc
9a2+b2+4c2−6ab+12ac−4bc=(3a)2+(−b)2+(2c)2+2(3a)(−b)+2(−b)(2c)+2(3a)(2c)=(3a−b+2c)2\begin{aligned} &9a^2 + b^2 + 4c^2 - 6ab + 12ac - 4bc \\ &= (3a)^2 + (-b)^2 + (2c)^2 + 2(3a)(-b) + 2(-b)(2c) + 2(3a)(2c) \\ &= (3a - b + 2c)^2 \end{aligned}
Answer

(i) (3a−b+2c)2(3a - b + 2c)^2

(ii) Factor 16s2+25t2−40st16s^2 + 25t^2 - 40st

Step 1 · Apply Identity for Square of a Binomial

Using the identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2:

  • a2=16s2  ⟹  a=4sa^2 = 16s^2 \implies a = 4s
  • b2=25t2  ⟹  b=5tb^2 = 25t^2 \implies b = 5t
  • Middle term: −2ab=−2(4s)(5t)=−40st-2ab = -2(4s)(5t) = -40st
16s2+25t2−40st=(4s)2−2(4s)(5t)+(5t)2=(4s−5t)2\begin{aligned} 16s^2 + 25t^2 - 40st &= (4s)^2 - 2(4s)(5t) + (5t)^2 \\ &= (4s - 5t)^2 \end{aligned}
Answer

(ii) (4s−5t)2(4s - 5t)^2

(iii) Factor r2−r−42r^2 - r - 42

Step 1 · Split the Middle Term

Find two numbers whose product is −42-42 and sum is −1-1. The numbers are −7-7 and 66.

r2−r−42=r2−7r+6r−42=r(r−7)+6(r−7)=(r−7)(r+6)\begin{aligned} r^2 - r - 42 &= r^2 - 7r + 6r - 42 \\ &= r(r - 7) + 6(r - 7) \\ &= (r - 7)(r + 6) \end{aligned}
Answer

(iii) (r−7)(r+6)(r - 7)(r + 6)

(iv) Factor 49g2+14gh+h249g^2 + 14gh + h^2

Step 1 · Apply Identity for Square of a Binomial

Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2:

  • a2=49g2  ⟹  a=7ga^2 = 49g^2 \implies a = 7g
  • b2=h2  ⟹  b=hb^2 = h^2 \implies b = h
  • Middle term: 2ab=2(7g)(h)=14gh2ab = 2(7g)(h) = 14gh
49g2+14gh+h2=(7g)2+2(7g)(h)+(h)2=(7g+h)2\begin{aligned} 49g^2 + 14gh + h^2 &= (7g)^2 + 2(7g)(h) + (h)^2 \\ &= (7g + h)^2 \end{aligned}
Answer

(iv) (7g+h)2(7g + h)^2

(v) Factor 64u2+121v2+4w2−176uv−32uw+44vw64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw

Step 1 · Apply Identity for Square of a Trinomial

Compare with (x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx:

  • x2=64u2  ⟹  x=8ux^2 = 64u^2 \implies x = 8u
  • y2=121v2y^2 = 121v^2 and 2xy=−176uv  ⟹  y=−11v2xy = -176uv \implies y = -11v
  • z2=4w2z^2 = 4w^2 and 2xz=−32uw  ⟹  z=−2w2xz = -32uw \implies z = -2w
  • Check: 2yz=2(−11v)(−2w)=44vw2yz = 2(-11v)(-2w) = 44vw
64u2+121v2+4w2−176uv−32uw+44vw=(8u)2+(−11v)2+(−2w)2+2(8u)(−11v)+2(−11v)(−2w)+2(8u)(−2w)=(8u−11v−2w)2\begin{aligned} &64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw \\ &= (8u)^2 + (-11v)^2 + (-2w)^2 + 2(8u)(-11v) + 2(-11v)(-2w) + 2(8u)(-2w) \\ &= (8u - 11v - 2w)^2 \end{aligned}
Answer

(v) (8u−11v−2w)2(8u - 11v - 2w)^2

Common Mistakes
  • Determining Signs in Trinomial Expansions: Look carefully at the negative cross-terms to identify which variables must carry the negative sign.
  • Middle Term Factor Signs: In quadratic splitting like r2−r−42r^2 - r - 42, choosing +7+7 and −6-6 gives +r+r instead of −r-r. Ensure the larger magnitude factor carries the sign of the middle term.
  • Equivalent Representations: Expressions such as (8u−11v−2w)2(8u - 11v - 2w)^2 and (−8u+11v+2w)2(-8u + 11v + 2w)^2 are algebraically equivalent since (−A)2=A2(-A)^2 = A^2.

More questions in Exercise 4.4

Q1

Fill in the blanks to complete the following identities:

Q2

Select and use the identity that will help you find the following products without multiplying directly:

(i) (41)2(41)^2

(ii) (27)2(27)^2

(iii) (23×17)(23 \times 17)

(iv) (135)2(135)^2

(v) (97)2(97)^2

(vi) (18×29)(18 \times 29)

(vii) (34×43)(34 \times 43)

(viii) (205)2(205)^2

Q3

Factor the following:

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

(ii) 16s2+25t2−40st16s^2 + 25t^2 - 40st

(iii) r2−r−42r^2 - r - 42

(iv) 49g2+14gh+h249g^2 + 14gh + h^2

(v) 64u2+121v2+4w2−176uv−32uw+44vw64u^2 + 121v^2 + 4w^2 - 176uv - 32uw + 44vw

← Back to Exploring Algebraic Identities