NCERT solution class 9 chapter 2 Polynomials exercise 2.5 mathematics

EXERCISE 2.5

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Q 1, Ex 2.5 - Polynomials - Chapter 2 - Maths Class 9th - NCERT

Q 1(i, ii & iii), Ex. 2.5, Page No 48, Polynomials, Maths Class 9th

Q 1(iv & v), Ex. 2.5, Page No 48, Polynomials, Class 9th NCERT Maths

Page No 43:

Question 1:

Determine which of the following polynomials has (x + 1) a factor:
(i) x³ + x² + x + 1
(ii) x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x + 1
(iv) x³-x²-(2+√2)x+√2

Answer:

(i) If (x + 1) is a factor of p(x) = x³ + x² + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).
p(x) = x³ + x² + x + 1
p(−1) = (−1)³ + (−1)² + (−1) + 1
= − 1 + 1 − 1 + 1 = 0
Hence, x + 1 is a factor of this polynomial.

(ii) If (x + 1) is a factor of p(x) = x⁴ + x³ + x² + x + 1, then p (−1) must be zero, otherwise (x + 1) is not a factor of p(x).
p(x) = x⁴ + x³ + x² + x + 1
p(−1) = (−1)⁴ + (−1)³ + (−1)² + (−1) + 1
= 1 − 1 + 1 −1 + 1 = 1
As p(− 1) ≠ 0,
Therefore, x + 1 is not a factor of this polynomial.

(iii) If (x + 1) is a factor of polynomial p(x) = x⁴ + 3x³ + 3x² + x + 1, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.
p(−1) = (−1)⁴ + 3(−1)³ + 3(−1)² + (−1) + 1
= 1 − 3 + 3 − 1 + 1 = 1
As p(−1) ≠ 0,
Therefore, x + 1 is not a factor of this polynomial.

(iv) If(x + 1) is a factor of polynomial p(x) =  x³-x²-(2+√2)x+√2, then p(−1) must be 0, otherwise (x + 1) is not a factor of this polynomial.
p(-1)=(-1)³-(-1)²-(2+√2)(-1)+√2
=-1-1+2+√2+√2
=2√2
As p(−1) ≠ 0,
Therefore, (x + 1) is not a factor of this polynomial.

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Q 2, Ex. 2.5, Page No 48, Polynomials, Class 9th Maths

Question 2:

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³ + x² − 2x − 1, g(x) = x + 1
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
(iii) p(x) = x³ − 4 x² + x + 6, g(x) = x − 3

Answer:

(i) If g(x) = x + 1 is a factor of the given polynomial p(x), then p(−1) must be zero.
p(x) = 2x³ + x² − 2x − 1
p(−1) = 2(−1)³ + (−1)² − 2(−1) − 1
= 2(−1) + 1 + 2 − 1 = 0
Hence, g(x) = x + 1 is a factor of the given polynomial.

(ii) If g(x) = x + 2 is a factor of the given polynomial p(x), then p(−2) must
be 0.
p(x) = x³ +3x² + 3x + 1
p(−2) = (−2)³ + 3(−2)² + 3(−2) + 1
= − 8 + 12 − 6 + 1
= −1
As p(−2) ≠ 0,
Hence, g(x) = x + 2 is not a factor of the given polynomial.

(iii) If g(x) = x − 3 is a factor of the given polynomial p(x), then p(3) must
be 0.
p(x) = x³ − 4 x² + x + 6
p(3) = (3)³ − 4(3)² + 3 + 6
= 27 − 36 + 9 = 0
Hence, g(x) = x − 3 is a factor of the given polynomial.

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Q 3, Ex. 2.5, Page No 48, Polynomials - NCERT Class 9th Maths Solutions

Page No 44:

Question 3:

Find the value of k, if x − 1 is a factor of p(x) in each of the following cases:
(i) p(x) = x² + x + k 
(ii) $p(x)=2 x^{2}+k x+\sqrt{2}$
(iii) $p(x)=k x^{2}-\sqrt{2} x+1$ 
(iv) p(x) = kx² − 3x + k

Answer:

If x − 1 is a factor of polynomial p(x), then p(1) must be 0.
(i) p(x) = x² + x + k
p(1) = 0
⇒ (1)² + 1 + k = 0
⇒ 2 + k = 0 ⇒ k = −2
Therefore, the value of k is −2.

(ii) $p(x)=2 x^{2}+k x+\sqrt{2}$
p(1) = 0
$\Rightarrow 2(1)^{2}+k(1)+\sqrt{2}=0$
$\Rightarrow 2+k+\sqrt{2}=0$
$\Rightarrow k=-2-\sqrt{2}=-(2+\sqrt{2})$

Therefore, the value of k is -(2+√2)

(iii) $p(x)=k x^{2}-\sqrt{2} x+1$
p(1) = 0
$\Rightarrow k(1)^{2}-\sqrt{2}(1)+1=0$
⇒k-√2+1=0
⇒$k=\sqrt{2}-1$

Therefore , the value of k is √2-1

(iv) p(x) = kx² − 3x + k
⇒ p(1) = 0
⇒ k(1)² − 3(1) + k = 0
⇒ k − 3 + k = 0
⇒ 2k − 3 = 0
⇒$k=\frac{3}{2}$
Therefore, the value of k is $\frac{3}{2}$ .

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Q 4, Ex 2.5 - Polynomials - Chapter 2 - Maths Class 9th - NCERT

Q 4, Ex. 2.5, Page No 49, Polynomials, Maths Class 9th

Question 4:

Factorise:
(i) 12x² − 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x − 6
(iv) 3x² − x − 4

Answer:

(i) 12x² − 7x + 1
We can find two numbers such that pq = 12 × 1 = 12 and p + q = −7. They are p = −4 and q = −3.
Here, 12x² − 7x + 1 = 12x² − 4x − 3x + 1
= 4x (3x − 1) − 1 (3x − 1)
= (3x − 1) (4x − 1)

(ii) 2x² + 7x + 3
We can find two numbers such that pq = 2 × 3 = 6 and p + q = 7.
They are p = 6 and q = 1.
Here, 2x² + 7x + 3 = 2x² + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
= (x + 3) (2x+ 1)

(iii) 6x² + 5x − 6
We can find two numbers such that pq = −36 and p + q = 5.
They are p = 9 and q = −4.
Here,
6x² + 5x − 6 = 6x² + 9x − 4x − 6
= 3x (2x + 3) − 2 (2x + 3)
= (2x + 3) (3x − 2)

(iv) 3x² − x − 4
We can find two numbers such that pq = 3 × (− 4) = −12
and p + q = −1.
They are p = −4 and q = 3.
Here,
3x² − x − 4 = 3x² − 4x + 3x − 4
= x (3x − 4) + 1 (3x − 4)
= (3x − 4) (x + 1)

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Q 5(i), Ex. 2.5, Page No 49, Polynomials, Maths Class 9th NCERT

Q 5(ii), Ex. 2.5, Page No 49, Polynomials, NCERT Maths Class 9th

Question 5:

Factorise:
(i) x³ − 2x² − x + 2
(ii) x³ + 3x² −9x − 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² − 2y − 1

Answer:

(i) Let p(x) = x³ − 2x² − x + 2
All the factors of 2 have to be considered. These are ± 1, ± 2.
By trial method,
p(−1) = (−1)³ − 2(−1)² − (−1) + 2
= −1 − 2 + 1 + 2 = 0
Therefore, (x +1 ) is factor of polynomial p(x).
Let us find the quotient on dividing x³ − 2x² − x + 2 by x + 1.
By long division,

It is known that,
Dividend = Divisor × Quotient + Remainder
∴ x³ − 2x² − x + 2 = (x + 1) (x² − 3x + 2) + 0
= (x + 1) [x² − 2x − x + 2]
= (x + 1) [x (x − 2) − 1 (x − 2)]
= (x + 1) (x − 1) (x − 2)
= (x − 2) (x − 1) (x + 1)

(ii) Let p(x) = x³ − 3x² − 9x − 5
All the factors of 5 have to be considered. These are ±1, ± 5.
By trial method,
p(−1) = (−1)³ − 3(−1)² − 9(−1) − 5
= − 1 − 3 + 9 − 5 = 0
Therefore, x + 1 is a factor of this polynomial.
Let us find the quotient on dividing x³ + 3x² − 9x − 5 by x + 1.
By long division,

It is known that,
Dividend = Divisor × Quotient + Remainder
∴ x³ − 3x² − 9x − 5 = (x + 1) (x² − 4x − 5) + 0
= (x + 1) (x² − 5x + x − 5)
= (x + 1) [(x (x − 5) +1 (x − 5)]
= (x + 1) (x − 5) (x + 1)
= (x − 5) (x + 1) (x + 1)

(iii) Let p(x) = x³ + 13x² + 32x + 20
All the factors of 20 have to be considered. Some of them are ±1,
± 2, ± 4, ± 5 ……
By trial method,
p(−1) = (−1)³ + 13(−1)² + 32(−1) + 20
= − 1 +13 − 32 + 20
= 33 − 33 = 0
As p(−1) is zero, therefore, x + 1 is a factor of this polynomial p(x).
Let us find the quotient on dividing x³ + 13x² + 32x + 20 by (x + 1).
By long division,

It is known that,
Dividend = Divisor × Quotient + Remainder
x³ + 13x² + 32x + 20 = (x + 1) (x² + 12x + 20) + 0
= (x + 1) (x² + 10x + 2x + 20)
= (x + 1) [x (x + 10) + 2 (x + 10)]
= (x + 1) (x + 10) (x + 2)
= (x + 1) (x + 2) (x + 10)
(iv) Let p(y) = 2y³ + y² − 2y − 1
By trial method,
p(1) = 2 ( 1)³ + (1)² − 2( 1) − 1
= 2 + 1 − 2 − 1= 0
Therefore, y − 1 is a factor of this polynomial.
Let us find the quotient on dividing 2y³ + y² − 2y − 1 by y ­− 1.

p(y) = 2y³ + y² − 2y − 1
= (y − 1) (2y² +3y + 1)
= (y − 1) (2y² +2y + y +1)
= (y − 1) [2y (y + 1) + 1 (y + 1)]
= (y − 1) (y + 1) (2y + 1)

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Question 5:

Factorise:
(i)  $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$

Answer 5:

It is known that,
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$
(i) $4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z-16 x z$
=(2x)²+(3y)²+(-4z)²+2(2x)(3y)+2(3y)(-4z)+2(2x)(-4z)
=(2x+3y-4z)²
=(2x+3y-4z)(2x+3y-4z)


(ii) $2 x^{2}+y^{2}+8 z^{2}-2 \sqrt{2} x y+4 \sqrt{2} y z-8 x z$
$=(-\sqrt{2} x)^{2}+(y)^{2}+(2 \sqrt{2} z)^{2}+2(-\sqrt{2} x)(y)+2(y)(2 \sqrt{2} z)+2(-\sqrt{2} x)(2 \sqrt{2} z)$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)^{2}$
$=(-\sqrt{2} x+y+2 \sqrt{2} z)(-\sqrt{2} x+y+2 \sqrt{2} z)$

Q 6 (i & ii), Ex. 2.5, Page No 49, Polynomials, Maths Class 9th

Q 6 (iii & iv), Ex. 2.5, Page No 49, Polynomials, Maths Class 9th

Question 6:

Write the following cubes in expanded form:
(i) $(2 x+1)^{3}$
(ii) $(2 a-3 b)^{3}$
(iii) $\left[\frac{3}{2} x+1\right]^{3}$
(iv) $\left[x-\frac{2}{3} y\right]^{3}$

Answer 6:

It is known that,
$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$
and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$

(i) $(2 x+1)^{3}=(2 x)^{3}+(1)^{3}$+3(2x)(1)(2x+1)
=8x³+1+6x(2x+1)
=8x³+1+12x²+6x
=8x³+12x²+6x+1


(ii) (2a-3b)³=(2a)³-(3b)³-3(2a)(3b)(2a-3b)
=8a³-27b³-18ab(2a-3b)
=8a³-27b³-36a²b+54ab²


(iii) $\left[\frac{3}{2} x+1\right]^{3}$
$=\left[\frac{3}{2} x\right]^{3}+(1)^{3}+3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x+1\right)$
$=\frac{27}{8} x^{3}+1+\frac{9}{2} x\left(\frac{3}{2} x+1\right)$
$=\frac{27}{8} x^{3}+1+\frac{27}{4} x^{2}+\frac{9}{2} x$
$=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1$

(vi) $\left[x-\frac{2}{3} y\right]^{3}$

$=x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$

$=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)$

$=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}$

Q 7, Ex. 2.5, Page No 49, Polynomials, NCERT Maths Class 9th

Question 7:

Evaluate the following using suitable identities:
(i) (99)³
(ii) (102)³
(iii) (998)³

Answer 7:

It is known that,
(a+b)³=a³+b³+3ab(a+b)
and
(a-b)³=a³-b³-3ab(a-b)

(i) (99)³ = (100 − 1)³
= (100)³ − (1)³ − 3(100) (1) (100 − 1)
= 1000000 − 1 − 300(99)
= 1000000 − 1 − 29700
= 970299

(ii) (102)³ = (100 + 2)³
= (100)³ + (2)³ + 3(100) (2) (100 + 2)
= 1000000 + 8 + 600 (102)
= 1000000 + 8 + 61200
= 1061208

(iii) (998)³= (1000 − 2)³
= (1000)³ − (2)³ − 3(1000) (2) (1000 − 2)
= 1000000000 − 8 − 6000(998)
= 1000000000 − 8 − 5988000
= 1000000000 − 5988008
= 994011992

Q 8(i, ii & iii), Ex. 2.5, Page No 49, Polynomials, Maths Class 9th

Q 8 (iv & v), Ex. 2.5, Page No 49, Polynomials, Maths Class 9th

Question 8:

Factorize each of the following:
(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
(iii) $27-125 a^{3}-135 a+225 a^{2}$
(iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$
(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$

Answer 8:

It is known that,
$(a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$
and $(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$

(i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$

$=(2 a)^{3}+(b)^{3}+3(2 a)^{2} b+3(2 a)(b)^{2}$

$=(2 a+b)^{3}$

=(2 a+b)(2 a+b)(2 a+b)


(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$

$=(2 a)^{3}-(b)^{3}-3(2 a)^{2} b+3(2 a)(b)^{2}$

$=(2 a-b)^{3}$

=(2 a-b)(2 a-b)(2 a-b)



(iii) $27-125 a^{3}-135 a+225 a^{2}$

=$(3)^{3}-(5 a)^{3}-3(3)^{2}(5 a)+3(3)(5 a)^{2}$

$=(3-5 a)^{3}$

=(3-5 a)(3-5 a)(3-5 a)


(iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$

$=(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}$

$=(4 a-3 b)^{3}$

=(4a-3b)(4a-3b)(4a-3b)


(v) $27 p^{3}-\frac{1}{216}-\frac{9}{2} p^{2}+\frac{1}{4} p$

$=(3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)^{2}\left(\frac{1}{6}\right)+3(3 p)\left(\frac{1}{6}\right)^{2}$

$=\left(3 p-\frac{1}{6}\right)^{3}$

$=\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$

Q 9, Ex. 2.5, Page No 49, Polynomials, Maths Class 9th

Question 9:

Verify:
(i) $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$
(ii) $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$

Answer 9:

(i) It is known that,
$(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$

$x^{3}+y^{3}=(x+y)^{3}-3 x y(x+y)$

$=(x+y)\left[(x+y)^{2}-3 x y\right]$

$=(x+y)\left(x^{2}+y^{2}+2 x y-3 x y\right)$

$=(x+y)\left(x^{2}+y^{2}-x y\right)$

$=(x+y)\left(x^{2}-x y+y^{2}\right)$


(ii) It is known that,
$(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)$

$x^{3}-y^{3}=(x-y)^{3}+3 x y(x-y)$

$=(x-y)\left[(x-y)^{2}+3 x y\right]$

$=(x-y)\left(x^{2}+y^{2}-2 x y+3 x y\right)$

$=(x-y)\left(x^{2}+y^{2}+x y\right)$

$=(x-y)\left(x^{2}+x y+y^{2}\right)$

Q 10, Ex. 2.5, Page No 49, Polynomials, Class 9th Maths

Question 10:

Factorise each of the following:
(i) $27 y^{3}+125 z^{3}$
(ii) $64 m^{3}-343 n^{3}$
[Hint: See question 9.]

Answer 10:

(i) $27 y^{3}+125 z^{3}$

$=(3 y)^{3}+(5 z)^{3}$

$=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right]$ $\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$

$=(3 y+5 z)\left[9 y^{2}+25 z^{2}-15 y z\right]$



(ii) $64 m^{3}-343 n^{3}$
 
$=(4 m)^{3}-(7 n)^{3}$

$=(4 m-7 n)\left[(4 m)^{2}+(7 n)^{2}+(4 m)(7 n)\right]$ $\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$

$=(4 m-7 n)\left[16 m^{2}+49 n^{2}+28 m n\right]$

Q 11, Ex. 2.5, Page No 49, Polynomials, Class 9th Maths

Question 11:

Factorise: $27 x^{3}+y^{3}+z^{3}-9 x y z$

Answer 11:

It is known that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

$\therefore 27 x^{3}+y^{3}+z^{3}-9 x y z$

$=(3 x)^{3}+(y)^{3}+(z)^{3}-3(3 x)(y)(z)$

$=(3 x+y+z)\left[(3 x)^{2}+y^{2}+z^{2}-(3 x)(y)-(y)(z)-z(3 x)\right]$

$=(3 x+y+z)\left[9 x^{2}+y^{2}+z^{2}-3 x y-y z-3 x z\right]$

Q 12, Ex. 2.5, Page No 49, Polynomials - NCERT Class 9th Maths Solutions

Question 12:

Verify that $x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$

Answer 12:

It is known that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$=\frac{1}{2}(x+y+z)\left[2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x\right]$
$=\frac{1}{2}(x+y+z)\left[\left(x^{2}+y^{2}-2 x y\right)+\left(y^{2}+z^{2}-2 y z\right)+\left(x^{2}+z^{2}-2 z x\right)\right]$
$=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right]$

Q 13, Ex. 2.5, Page No 49, Polynomials - NCERT Class 9th Maths Solutions

Question 13:

If x + y + z = 0, show that $x^3 + y^3 + z^3 = 3xyz$.

Answer 13:

It is known that,
$x^{3}+y^{3}+z^{3}-3 x y z$
=(x+y+z)$\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$

Put x + y + z = 0,
$x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$x^{3}+y^{3}+z^{3}-3 x y z=0$
$x^{3}+y^{3}+z^{3}=3 x y z$

Q 14, Ex. 2.5, Page No 49, Polynomials - NCERT Class 9th Maths Solutions

Question 14:

Without actually calculating the cubes, find the value of each of the following:
(i) $(-12)^{3}+(7)^{3}+(5)^{3}$
(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$

Answer 14:

(i) $(-12)^{3}+(7)^{3}+(5)^{3}$
Let x = −12, y = 7, and z = 5
It can be observed that,
x + y + z = − 12 + 7 + 5 = 0
It is known that if x + y + z = 0, then
$x^{3}+y^{3}+z^{3}=3 x y z$
∴ $(-12)^{3}+(7)^{3}+(5)^{3}=3(-12)(7)(5)$
= −1260

(ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$
Let x = 28, y = −15, and z = −13
It can be observed that,
x + y + z = 28 + (−15) + (−13) = 28 − 28 = 0
It is known that if x + y + z = 0, then
$x^{3}+y^{3}+z^{3}=3 x y z$

$\therefore(28)^{3}+(-15)^{3}+(-13)^{3}=3(28)(-15)(-13)$
=16380

Q 15, Ex 2.5, Page No 49, Polynomials - NCERT Class 9th Maths Solutions

Question 15:

Give possible expressions for the length and breadth of each of thefollowing rectangles, in which their areas are given:

Answer 15:

Area = Length × Breadth
The expression given for the area of the rectangle has to be factorised. One of its factors will be its length and the other will be its breadth.
(i) $25 a^{2}-35 a+12=25 a^{2}-15 a-20 a+12$
=5a(5a-3)-4(5a-3)
=(5a-3)(5a-4)
Therefore, possible length = 5a − 3
And, possible breadth = 5a − 4

(ii) $35 y^{2}+13 y-12=35 y^{2}+28 y-15 y-12$
=7y(5y+4)-3(5y+4)
=(5y+4)(7y-3)
Therefore, possible length = 5y + 4
And, possible breadth = 7y − 3

Q 16, Ex. 2.5, Page No 50, Polynomials - NCERT Class 9th Maths Solutions