NCERT Solutions Class 9 Maths Exercise 4.4 — Exploring Algebraic Identities

NCERT Class 9 Mathematics — Ganita Manjari Part I, Exploring Algebraic Identities, Exercise Set 4.4. All 3 questions solved.

Not every quadratic is a perfect square. This exercise introduces the general pattern

$$x^2 + (a+b)x + ab = (x + a)(x + b)$$

so factorising $x^2 + px + q$ means finding two numbers that multiply to $q$ and add to $p$. Start from the factor pairs of $q$ — there are only finitely many — and test each against the sum.

Two more identities join the list here:

$$a^2 – b^2 = (a+b)(a-b) \qquad\qquad a^2 = (a+b)(a-b) + b^2$$

the second of which is Śhrīdharāchārya’s method for squaring numbers quickly.

Key insight. With a leading coefficient bigger than $1$, as in $10x^2 – 11x – 6$, look for two numbers multiplying to (leading coefficient) $\times$ (constant) and adding to the middle coefficient — here $10 \times (-6) = -60$ and $-11$, giving $-15$ and $4$. Then split the middle term and group. It is the same search, over a different product.

Question 1

Fill in the blanks to complete the following identities:

Solution.

(i) $s^2 – 11s + 24 = (\underline{\phantom{xxxx}})(\underline{\phantom{xxxx}})$

Two numbers multiplying to $24$ and adding to $-11$. Both must be negative (their product is positive, their sum negative). The factor pairs of $24$ are $1 \times 24$, $2 \times 12$, $3 \times 8$, $4 \times 6$ — and $-3$ and $-8$ add to $-11$ ✓

$$s^2 – 11s + 24 = (s – 3)(s – 8)$$

(ii) $(\underline{\phantom{xxxx}})(x + 1) = 3x^2 – 4x – 7$

One factor is known, so divide — or find the other factor by inspection. The leading term $3x^2$ must come from $3x \times x$, and the constant $-7$ from $-7 \times 1$:

$$(3x – 7)(x + 1) = 3x^2 + 3x – 7x – 7 = 3x^2 – 4x – 7 \; \checkmark$$

(iii) $10x^2 – 11x – 6 = (2x – \underline{\phantom{x}})(\underline{\phantom{x}} + 2)$

The shape of the answer is given. The second bracket must start with a term multiplying $2x$ to give $10x^2$, so it is $5x$. Then the constants must multiply to $-6$: with $+2$ in the second bracket, the first needs $-3$.

$$(2x – 3)(5x + 2) = 10x^2 + 4x – 15x – 6 = 10x^2 – 11x – 6 \; \checkmark$$

(iv) $6x^2 + 7x + 2 = (\underline{\phantom{xxxx}})(\underline{\phantom{xxxx}})$

Split the middle term. Two numbers multiplying to $6 \times 2 = 12$ and adding to $7$: these are $3$ and $4$.

$$6x^2 + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (2x + 1)(3x + 2)$$

Question 2

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

Solution. Three tools are available: $(a+b)^2$, $(a-b)^2$ and $a^2 – b^2 = (a+b)(a-b)$. The last is for products of two different numbers that sit symmetrically about a round one.

(i) $(41)^2$

$(40+1)^2 = 1600 + 80 + 1 = 1681$

(ii) $(27)^2$

$(30-3)^2 = 900 – 180 + 9 = 729$

(iii) $23 \times 17$

These average $20$, sitting $3$ either side — so use $a^2 – b^2$ with $a = 20$, $b = 3$:

$$(20+3)(20-3) = 400 – 9 = 391$$

(iv) $(135)^2$

$(140-5)^2 = 19600 – 1400 + 25 = 18225$

(Or use Śhrīdharāchārya’s form: $135^2 = (135+5)(135-5) + 5^2 = 140 \times 130 + 25 = 18200 + 25 = 18225$.)

(v) $(97)^2$

$(100-3)^2 = 10000 – 600 + 9 = 9409$

(vi) $18 \times 29$

These are not symmetric about a round number, so $a^2 – b^2$ does not apply directly. Use the $(x+a)(x+b)$ pattern instead, with $x = 20$:

$$(20 – 2)(20 + 9) = 400 + 20(9 – 2) – 18 = 400 + 140 – 18 = 522$$

(vii) $34 \times 43$

Again not symmetric. Take $x = 40$:

$$(40 – 6)(40 + 3) = 1600 + 40(3 – 6) – 18 = 1600 – 120 – 18 = 1462$$

(viii) $(205)^2$

$(200+5)^2 = 40000 + 2000 + 25 = 42025$

Question 3

Factor the following:

Solution.

(i) $9a^2 + b^2 + 4c^2 – 6ab + 12ac – 4bc$

Six terms with three squares — the three-term identity. The squares are $(3a)^2$, $b^2$, $(2c)^2$, and the signs of the cross terms say which letters carry a minus. Try $3a$, $-b$, $2c$:

$$2(3a)(-b) = -6ab \; \checkmark \qquad 2(-b)(2c) = -4bc \; \checkmark \qquad 2(2c)(3a) = 12ac \; \checkmark$$

$$= (3a – b + 2c)^2$$

(ii) $16s^2 + 25t^2 – 40st$

Rearranged, this is $16s^2 – 40st + 25t^2$: a three-term square with $a = 4s$, $b = 5t$ and $2(4s)(5t) = 40st$ ✓

$$= (4s – 5t)^2$$

(iii) $r^2 – r – 42$

Two numbers multiplying to $-42$ and adding to $-1$. Opposite signs, and the pairs of $42$ are $1\times42$, $2\times21$, $3\times14$, $6\times7$ — so $-7$ and $6$:

$$r^2 – 7r + 6r – 42 = r(r – 7) + 6(r – 7) = (r – 7)(r + 6)$$

(iv) $49g^2 + 14gh + h^2$

$a = 7g$, $b = h$, and $2(7g)(h) = 14gh$ ✓

$$= (7g + h)^2$$

(v) $64u^2 + 121v^2 + 4w^2 – 176uv – 32uw + 44vw$

Squares $(8u)^2$, $(11v)^2$, $(2w)^2$. Two cross terms are negative and one positive, and both negatives involve $u$ — so $u$ is the odd one out. Try $8u$, $-11v$, $-2w$:

$$2(8u)(-11v) = -176uv \; \checkmark$$ $$2(-11v)(-2w) = +44vw \; \checkmark$$ $$2(-2w)(8u) = -32uw \; \checkmark$$

$$= (8u – 11v – 2w)^2$$

Common mistakes

• Question 1(i), choosing $3$ and $8$ instead of $-3$ and $-8$. They multiply to $24$ but add to $+11$, giving $s^2 + 11s + 24$. With a positive constant and negative middle term, both numbers are negative.
• Question 1(iv), looking for factors of $2$ alone. With a leading coefficient of $6$, the product to aim for is $6 \times 2 = 12$, not $2$. Using $1$ and $2$ gives no valid split.
• Question 2(iii), squaring instead. $23 \times 17$ is a product of two different numbers, so no square identity applies. They lie $3$ either side of $20$, which is the signature of $a^2 – b^2$.
• Question 2(vi) and (vii), forcing $a^2 – b^2$. $18$ and $29$ do not sit symmetrically about any round number, so that identity cannot be used. The general $(x+a)(x+b)$ pattern can.
• Question 3(i) and (v), guessing the signs. Work them out from the cross terms: a letter appearing in two negative cross terms and one positive is the one carrying the minus sign. Guessing and expanding to check is slower but also works.
• Question 3(ii), giving up because the terms are out of order. $16s^2 + 25t^2 – 40st$ is the same expression as $16s^2 – 40st + 25t^2$. Reordering into the standard shape is always the first step.
• Question 3(iii), treating it as a perfect square. $r^2 – r – 42$ has a negative constant term, and no perfect square does. It must split into two different brackets.

Practise next

• Exercise Set 4.5 — simplifying rational expressions, where every numerator and denominator must first be factorised by the methods used here.
• The End-of-Chapter Exercises, Questions 5, 6 and 9 — factorising to recover the dimensions of a rectangle, a cuboid and a pool, which is this exercise applied to a situation.