Chapter 2: Polynomials – Notes, Practice & Exercise Solutions (Class 9, NCERT/CBSE)
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1-Mark Questions (20)
1) Define a polynomial in one variable.
An expression of the form \(a_n x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\) where \(a_i\in \mathbb{R}\) and \(a_n\ne 0\), with all exponents non-negative integers.
2) What is the degree of \(7-3x+5x^4\)?
\(4\).
3) Is \(x+\dfrac{2}{x}\) a polynomial?
No, because the exponent of \(x\) in \(\dfrac{2}{x}=2x^{-1}\) is negative.
4) Identify monomial, binomial or trinomial: \(5x^3-2\).
Binomial (two terms).
5) Give the multiplicative identity in \(\mathbb{R}\).
\(1\).
6) Zero of a constant non-zero polynomial?
None.
7) Find the coefficient of \(x^2\) in \(2x^2-3x+1\).
\(2\).
8) State Factor Theorem in one line.
\((x-a)\) is a factor of \(p(x)\) iff \(p(a)=0\).
9) Write the identity: \((x+y)^3=\ ?\)
\((x+y)^3=x^3+3x^2y+3xy^2+y^3\).
10) Degree of the zero polynomial?
Not defined.
11) Is every integer a polynomial value?
Yes, integers are constant polynomials (degree 0, except 0).
12) Number of zeros of a linear polynomial.
Exactly one.
13) Write \((x-y)^2\).
\(x^2-2xy+y^2\).
14) Classify \(3t^2+t\).
Polynomial in one variable \(t\); degree \(2\) (quadratic).
15) Highest possible number of terms in a cubic in one variable.
At most \(4\) terms: \(ax^3+bx^2+cx+d\).
16) If \(p(x)=2x+5\), find its zero.
Solve \(2x+5=0\Rightarrow x=-\dfrac{5}{2}\).
17) Write the identity \(x^2-y^2=\ ?\)
\((x+y)(x-y)\).
18) Is \(x^{1/2}+3\) a polynomial?
No; exponent \(1/2\) is not a whole number.
19) Find the constant term in \(5x^3-2x+7\).
\(7\).
20) Evaluate \(p(0)\) for \(p(x)=x^2-3x+2\).
\(2\).
2-Mark Questions (20)
1) Classify as terminating/non-terminating repeating: \( \dfrac{7}{40},\ \dfrac{5}{12}\).
\(\tfrac{7}{40}\) terminates (denominator \(=2^3\cdot 5\)). \(\tfrac{5}{12}\) repeats (denominator has prime \(3\)).
2) Find three rational numbers between \(\dfrac{1}{4}\) and \(\dfrac{1}{3}\).
As decimals: \(0.25\) and \(0.333\ldots\); choose \(0.26,0.28,0.3\Rightarrow \tfrac{13}{50},\tfrac{7}{25},\tfrac{3}{10}\).
3) Determine whether \(x+1\) is a factor of \(x^3+x^2+x+1\).
Check \(x=-1\): \((-1)^3+(-1)^2+(-1)+1=-1+1-1+1=0\). Yes, factor.
4) Factorise \(12x^2-7x+1\).
\((3x-1)(4x-1)\).
5) Convert \(3.0\overline{6}\) to a fraction in lowest form.
Let \(x=3.0\overline{6}\). Then \(10x=30.\overline{6},\, 100x=306.\overline{6}\). Subtract: \(90x=276\Rightarrow x=\dfrac{276}{90}=\dfrac{46}{15}\).
6) Using identities, expand \((x+2y+4z)^2\).
\(x^2+4y^2+16z^2+4xy+8xz+16yz\).
7) Evaluate \(103\times 107\) without long multiplication.
\((100+3)(100+7)=10000+1000+700+21=11021\).
8) Find the zero of \(p(x)=3x-2\).
\(x=\dfrac{2}{3}\).
9) Decide if \(x^2-\dfrac{1}{100}y^2\) is factorable via difference of squares.
Yes: \(\left(x-\dfrac{y}{10}\right)\left(x+\dfrac{y}{10}\right)\).
10) If \(p(x)=ax+b\ (a\ne 0)\), prove it has exactly one zero.
Solve \(ax+b=0\Rightarrow x=-\dfrac{b}{a}\) (unique).
11) Factorise \(6x^2+5x-6\).
\((3x-2)(2x+3)\).
12) Check if \(x-3\) divides \(x^3-4x^2+x+6\).
\(p(3)=27-36+3+6=0\Rightarrow\) yes.
13) Write \((3a-4b)^2\).
\(9a^2-24ab+16b^2\).
14) Evaluate \(95\times 96\).
\((100-5)(100-4)=10000-900+20=9120\).
15) Factorise \(2x^2+7x+3\).
\((2x+1)(x+3)\).
16) Express \(x^3+y^3\) as a product.
\((x+y)(x^2-xy+y^2)\).
17) If \(p(x)=(x+1)(x-2)\), find all its zeros.
\(x=-1,\,2\).
18) Expand \((2a-3b)^3\).
\(8a^3-36a^2b+54ab^2-27b^3\).
19) Find the degree and type of \(5-7y+y^3\).
Degree \(3\), cubic trinomial.
20) Determine if \(x+1\) is a factor of \(x^4+x^3+x^2+x+1\).
Check \(x=-1\): \(1-1+1-1+1=1\ne 0\Rightarrow\) not a factor.
3-Mark Questions (20)
1) Prove: a rational number \(\dfrac{p}{q}\) (lowest form) has a terminating decimal iff \(q=2^m5^n\) for some \(m,n\in \mathbb{Z}_{\ge 0}\).
(\(\Rightarrow\)) If it terminates at \(k\) places, \(\frac{p}{q}=\frac{N}{10^k}=\frac{N}{2^k5^k}\Rightarrow q\mid 2^k5^k\). (\(\Leftarrow\)) If \(q=2^m5^n\), multiply numerator/denominator suitably to get denominator \(10^{\max(m,n)}\), hence terminating.
2) Factorise \(x^3-2x^2-x+2\).
Group: \(x^2(x-2)-1(x-2)=(x-2)(x^2-1)=(x-2)(x-1)(x+1)\).
3) If \(x+y+z=0\), show \(x^3+y^3+z^3=3xyz\).
Use identity \(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\). LHS \(=0\Rightarrow x^3+y^3+z^3=3xyz\).
4) Factorise \(x^3-3x^2-9x-5\).
Trial: \(x=5\) gives 0 \(\Rightarrow (x-5)(x^2+2x+1)=(x-5)(x+1)^2\).
5) Using identities, expand \((3a+4b)^3\).
\(27a^3+108a^2b+144ab^2+64b^3\).
6) Factorise \(4x^2+9y^2+16z^2+12xy-24yz-16xz\).
(Perfect square): \((2x+3y-4z)^2\).
7) Find all zeros of \(2y^3+y^2-2y-1\).
\(y=1\) is a zero. Divide: \(2y^2+3y+1=(2y+1)(y+1)\). Zeros: \(1,-\tfrac{1}{2},-1\).
8) Show that \((\sqrt{2}x-y-2\sqrt{2}z)^2=2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz\).
Expand \((A+B+C)^2\) with \(A=\sqrt{2}x,\ B=-y,\ C=-2\sqrt{2}z\).
9) Evaluate \(998^3\) using identities.
\((1000-2)^3=10^9-3\cdot10^6\cdot2+3\cdot1000\cdot4-8=994,011,992\).
10) If \(x-1\) is a factor of \(kx^2-3x+k\), find \(k\).
Put \(x=1\): \(k-3+k=0\Rightarrow 2k=3\Rightarrow k=\tfrac{3}{2}\).
11) Convert \(2.4\overline{07}\) to a rational number.
Let \(x=2.4\overline{07}\). Then \(10x=24.\overline{07},\,1000x=2407.\overline{07}\). Subtract: \(990x=2383\Rightarrow x=\dfrac{2383}{990}\).
12) Prove \((x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx\).
Expand using \((a+b)^2\) twice.
13) Factorise \(8a^3-b^3-12a^2b+6ab^2\).
Recognise cube: \((2a-b)^3\).
14) Show that between any two distinct reals there is a rational number.
For \(a
15) Factorise \(x^3+13x^2+32x+20\).
\(x=-1\) is a root \(\Rightarrow (x+1)(x^2+12x+20)=(x+1)(x+2)(x+10)\).
16) Expand \((3a-7b-c)^2\).
\(9a^2+49b^2+c^2-42ab-6ac+14bc\).
17) Evaluate \(104\times 96\) using identities.
\((100+4)(100-4)=10000-16=9984\).
18) Factorise \(27-125a^3-135a+225a^2\).
It is \((3-5a)^3\).
19) Factorise \(64a^3-27b^3-144a^2b+108ab^2\).
It is \((4a-3b)^3\).
20) Verify \(x^3-y^3=(x-y)(x^2+xy+y^2)\).
Multiply RHS and collect like terms to get \(x^3-y^3\).
Exercise 2.1 – Solutions
1) Which are polynomials in one variable?
(i) \(4x^2-3x+7\): Yes (one variable \(x\), whole-number exponents).
(ii) \(y^2+2\): Yes (one variable \(y\)).
(iii) \(3t^2+t\): Yes (one variable \(t\)).
(iv) \(y+\dfrac{2}{y}\): No (term \(y^{-1}\)).
(v) \(x^{10}+y^3+t^{50}\): No (more than one variable).
(ii) \(y^2+2\): Yes (one variable \(y\)).
(iii) \(3t^2+t\): Yes (one variable \(t\)).
(iv) \(y+\dfrac{2}{y}\): No (term \(y^{-1}\)).
(v) \(x^{10}+y^3+t^{50}\): No (more than one variable).
2) Write the coefficient of \(x^2\) in each: (i) \(2+x^2+x\) (ii) \(2-x^2+x^3\) (iii) \(2x^2+\pi x\) (iv) \(2-\dfrac{1}{x}\)
(i) \(1\) (ii) \(-1\) (iii) \(2\) (iv) \(0\).
3) One example each: (a) Binomial of degree \(35\) (b) Monomial of degree \(100\).
(a) \(x^{35}+1\) (b) \(7y^{100}\).
4) Degree of: (i) \(5x^3+4x^2+7x\) (ii) \(4-y^2\) (iii) \(5t-7\) (iv) \(3\)
(i) \(3\) (ii) \(2\) (iii) \(1\) (iv) \(0\).
5) Classify as linear/quadratic/cubic: (i) \(x^2+x\) (ii) \(x-x^3\) (iii) \(y+y^2+4\) (iv) \(1+x\) (v) \(3t\) (vi) \(r^2\) (vii) \(7x^3\)
(i) Quadratic (ii) Cubic (iii) Quadratic (iv) Linear (v) Linear (vi) Quadratic (vii) Cubic.
Exercise 2.2 – Solutions
1) Find \(5x-4x^2+3\) at (i) \(x=0\) (ii) \(x=-1\) (iii) \(x=2\).
(i) \(3\) (ii) \(-6\) (iii) \(-3\).
2) Find \(p(0),p(1),p(2)\): (i) \(y^2-y+1\) (ii) \(2+t+2t^2-t^3\) (iii) \(x^3\) (iv) \((x-1)(x+1)\)
(i) \(1,\,1,\,3\) (ii) \(2,\,4,\,4\) (iii) \(0,\,1,\,8\) (iv) \(-1,\,0,\,3\).
3) Verify zeros:
(i) \(p(x)=3x+1,\ x=-\tfrac13\): \(0\) ✓
(ii) \(p(x)=5x-\pi,\ x=\tfrac45\): \(4-\pi\ne 0\) ✗
(iii) \(p(x)=x^2-1,\ x=1,-1\): both \(0\) ✓
(iv) \(p(x)=(x+1)(x-2),\ x=-1,2\): both \(0\) ✓
(v) \(p(x)=x^2,\ x=0\): \(0\) ✓
(vi) \(p(x)=\ell x+m,\ x=-\tfrac{m}{\ell}\): \(0\) ✓
(vii) \(p(x)=3x^2-1,\ x=\pm \tfrac{1}{\sqrt3}\): \(0\) ✓
(viii) \(p(x)=2x+1,\ x=\tfrac12\): \(2\ne 0\) ✗
(ii) \(p(x)=5x-\pi,\ x=\tfrac45\): \(4-\pi\ne 0\) ✗
(iii) \(p(x)=x^2-1,\ x=1,-1\): both \(0\) ✓
(iv) \(p(x)=(x+1)(x-2),\ x=-1,2\): both \(0\) ✓
(v) \(p(x)=x^2,\ x=0\): \(0\) ✓
(vi) \(p(x)=\ell x+m,\ x=-\tfrac{m}{\ell}\): \(0\) ✓
(vii) \(p(x)=3x^2-1,\ x=\pm \tfrac{1}{\sqrt3}\): \(0\) ✓
(viii) \(p(x)=2x+1,\ x=\tfrac12\): \(2\ne 0\) ✗
4) Zero of each polynomial:
(i) \(x+5\Rightarrow -5\) (ii) \(x-5\Rightarrow 5\) (iii) \(2x+5\Rightarrow -\tfrac{5}{2}\)
(iv) \(3x-2\Rightarrow \tfrac{2}{3}\) (v) \(3x\Rightarrow 0\) (vi) \(ax\ (a\ne 0)\Rightarrow 0\) (vii) \(cx+d\Rightarrow -\tfrac{d}{c}\).
(iv) \(3x-2\Rightarrow \tfrac{2}{3}\) (v) \(3x\Rightarrow 0\) (vi) \(ax\ (a\ne 0)\Rightarrow 0\) (vii) \(cx+d\Rightarrow -\tfrac{d}{c}\).
Exercise 2.3 – Solutions
1) Determine whether \((x+1)\) is a factor of:
(i) \(x^3+x^2+x+1\): \(p(-1)=0\Rightarrow\) Yes.
(ii) \(x^4+x^3+x^2+x+1\): \(p(-1)=1\Rightarrow\) No.
(iii) \(x^4+3x^3+3x^2+x+1\): \(p(-1)=1\Rightarrow\) No.
(ii) \(x^4+x^3+x^2+x+1\): \(p(-1)=1\Rightarrow\) No.
(iii) \(x^4+3x^3+3x^2+x+1\): \(p(-1)=1\Rightarrow\) No.
2) Use Factor Theorem: is \(g(x)\) a factor of \(p(x)\)?
(i) \(p=2x^3+x^2-2x-1,\ g=x+1\): \(p(-1)=0\Rightarrow\) Yes.
(ii) \(p=x^3+3x^2+3x+1,\ g=x+2\): \(p(-2)=-1\Rightarrow\) No.
(iii) \(p=x^3-4x^2+x+6,\ g=x-3\): \(p(3)=0\Rightarrow\) Yes.
(ii) \(p=x^3+3x^2+3x+1,\ g=x+2\): \(p(-2)=-1\Rightarrow\) No.
(iii) \(p=x^3-4x^2+x+6,\ g=x-3\): \(p(3)=0\Rightarrow\) Yes.
3) Find \(k\) if \(x-1\) is a factor of \(p(x)\):
(i) \(x^2+x+k:\ k=-2\). (ii) \(2x^2+kx+2:\ k=-4\). (iii) \(kx^2-2x+1:\ k=1\). (iv) \(kx^2-3x+k:\ k=\tfrac{3}{2}\).
4) Factorise:
(i) \(12x^2-7x+1=(3x-1)(4x-1)\).
(ii) \(2x^2+7x+3=(2x+1)(x+3)\).
(iii) \(6x^2+5x-6=(3x-2)(2x+3)\).
(iv) \(3x^2-x-4=(3x-4)(x+1)\).
(ii) \(2x^2+7x+3=(2x+1)(x+3)\).
(iii) \(6x^2+5x-6=(3x-2)(2x+3)\).
(iv) \(3x^2-x-4=(3x-4)(x+1)\).
5) Factorise:
(i) \(x^3-2x^2-x+2=(x-2)(x-1)(x+1)\).
(ii) \(x^3-3x^2-9x-5=(x-5)(x+1)^2\).
(iii) \(x^3+13x^2+32x+20=(x+1)(x+2)(x+10)\).
(iv) \(2y^3+y^2-2y-1=(y-1)(2y+1)(y+1)\).
(ii) \(x^3-3x^2-9x-5=(x-5)(x+1)^2\).
(iii) \(x^3+13x^2+32x+20=(x+1)(x+2)(x+10)\).
(iv) \(2y^3+y^2-2y-1=(y-1)(2y+1)(y+1)\).
Exercise 2.4 – Solutions
1) Use identities to find products:
(i) \((x+4)(x+10)=x^2+14x+40\).
(ii) \((x+8)(x-10)=x^2-2x-80\).
(iii) \((3x+4)(3x-5)=9x^2-3x-20\).
(iv) \(\left(y^2+\tfrac{3}{2}\right)\left(y^2-\tfrac{3}{2}\right)=y^4-\tfrac{9}{4}\).
(v) \((3-2x)(3+2x)=9-4x^2\).
(ii) \((x+8)(x-10)=x^2-2x-80\).
(iii) \((3x+4)(3x-5)=9x^2-3x-20\).
(iv) \(\left(y^2+\tfrac{3}{2}\right)\left(y^2-\tfrac{3}{2}\right)=y^4-\tfrac{9}{4}\).
(v) \((3-2x)(3+2x)=9-4x^2\).
2) Evaluate without direct multiplication:
(i) \(103\times 107=11021\). (ii) \(95\times 96=9120\). (iii) \(104\times 96=9984\).
3) Factorise using identities:
(i) \(9x^2+6xy+y^2=(3x+y)^2\).
(ii) \(4y^2-4y+1=(2y-1)^2\).
(iii) \(x^2-\dfrac{1}{100}y^2=\left(x-\dfrac{y}{10}\right)\left(x+\dfrac{y}{10}\right)\).
(ii) \(4y^2-4y+1=(2y-1)^2\).
(iii) \(x^2-\dfrac{1}{100}y^2=\left(x-\dfrac{y}{10}\right)\left(x+\dfrac{y}{10}\right)\).
4) Expand:
(i) \((x+2y+4z)^2=x^2+4y^2+16z^2+4xy+8xz+16yz\).
(ii) \((2x-y+z)^2=4x^2+y^2+z^2-4xy+4xz-2yz\).
(iii) \((-2x+3y+2z)^2=4x^2+9y^2+4z^2-12xy-8xz+12yz\).
(iv) \((3a-7b-c)^2=9a^2+49b^2+c^2-42ab-6ac+14bc\).
(v) \((-2x+5y-3z)^2=4x^2+25y^2+9z^2-20xy+12xz-30yz\).
(vi) \(\left(\dfrac{a}{4}-\dfrac{b}{2}\right)^2=\dfrac{a^2}{16}+\dfrac{b^2}{4}-\dfrac{ab}{4}\).
(ii) \((2x-y+z)^2=4x^2+y^2+z^2-4xy+4xz-2yz\).
(iii) \((-2x+3y+2z)^2=4x^2+9y^2+4z^2-12xy-8xz+12yz\).
(iv) \((3a-7b-c)^2=9a^2+49b^2+c^2-42ab-6ac+14bc\).
(v) \((-2x+5y-3z)^2=4x^2+25y^2+9z^2-20xy+12xz-30yz\).
(vi) \(\left(\dfrac{a}{4}-\dfrac{b}{2}\right)^2=\dfrac{a^2}{16}+\dfrac{b^2}{4}-\dfrac{ab}{4}\).
5) Factorise (perfect squares):
(i) \(4x^2+9y^2+16z^2+12xy-24yz-16xz=(2x+3y-4z)^2\).
(ii) \(2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz=(\sqrt{2}x-y-2\sqrt{2}z)^2\).
(ii) \(2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz=(\sqrt{2}x-y-2\sqrt{2}z)^2\).
6) Write cubes in expanded form:
(i) \((2x+1)^3=8x^3+12x^2+6x+1\).
(ii) \((2a-3b)^3=8a^3-36a^2b+54ab^2-27b^3\).
(iii) \(\left(\dfrac{x}{2}+\dfrac{1}{3}\right)^3=\dfrac{x^3}{8}+\dfrac{1}{27}+\dfrac{x^2}{4}+\dfrac{x}{6}\).
(iv) \(\left(\dfrac{2x}{3}-y\right)^3=\dfrac{8x^3}{27}-\dfrac{4x^2y}{3}+2xy^2-y^3\).
(ii) \((2a-3b)^3=8a^3-36a^2b+54ab^2-27b^3\).
(iii) \(\left(\dfrac{x}{2}+\dfrac{1}{3}\right)^3=\dfrac{x^3}{8}+\dfrac{1}{27}+\dfrac{x^2}{4}+\dfrac{x}{6}\).
(iv) \(\left(\dfrac{2x}{3}-y\right)^3=\dfrac{8x^3}{27}-\dfrac{4x^2y}{3}+2xy^2-y^3\).
7) Evaluate using identities:
(i) \(99^3=(100-1)^3=970,299\).
(ii) \(102^3=(100+2)^3=1,061,208\).
(iii) \(998^3=(1000-2)^3=994,011,992\).
(ii) \(102^3=(100+2)^3=1,061,208\).
(iii) \(998^3=(1000-2)^3=994,011,992\).
8) Factorise each:
(i) \(8a^3+b^3+12a^2b+6ab^2=(2a+b)^3\).
(ii) \(8a^3-b^3-12a^2b+6ab^2=(2a-b)^3\).
(iii) \(27-125a^3-135a+225a^2=(3-5a)^3\).
(iv) \(64a^3-27b^3-144a^2b+108ab^2=(4a-3b)^3\).
(v) \(27p^3-\dfrac{1}{216}-\dfrac{9}{2}p^2+\dfrac{1}{4}p=\left(3p-\dfrac{1}{6}\right)^3\).
(ii) \(8a^3-b^3-12a^2b+6ab^2=(2a-b)^3\).
(iii) \(27-125a^3-135a+225a^2=(3-5a)^3\).
(iv) \(64a^3-27b^3-144a^2b+108ab^2=(4a-3b)^3\).
(v) \(27p^3-\dfrac{1}{216}-\dfrac{9}{2}p^2+\dfrac{1}{4}p=\left(3p-\dfrac{1}{6}\right)^3\).
9) Verify identities:
(i) \(x^3+y^3=(x+y)(x^2-xy+y^2)\).
(ii) \(x^3-y^3=(x-y)(x^2+xy+y^2)\).
(ii) \(x^3-y^3=(x-y)(x^2+xy+y^2)\).
10) Factorise: (i) \(27y^3+125z^3\) (ii) \(64m^3-343n^3\).
(i) \((3y+5z)(9y^2-15yz+25z^2)\).
(ii) \((4m-7n)(16m^2+28mn+49n^2)\).
(ii) \((4m-7n)(16m^2+28mn+49n^2)\).
11) Factorise \(27x^3+y^3+z^3-9xyz\).
Let \(X=3x\). Then \(X^3+y^3+z^3-3Xyz=(X+y+z)(X^2+y^2+z^2-XY-YZ-ZX)=(3x+y+z)(9x^2+y^2+z^2-3xy-3xz-yz)\).
12) Verify: \(x^3+y^3+z^3-3xyz=\dfrac12(x+y+z)\big[(x-y)^2+(y-z)^2+(z-x)^2\big]\).
Start from RHS, expand squares and simplify; it equals LHS.
13) If \(x+y+z=0\), show \(x^3+y^3+z^3=3xyz\).
From Q12, RHS becomes 0; hence LHS \(=3xyz\).
14) Without calculating cubes, evaluate: (i) \((-12)^3+7^3+5^3\) (ii) \(28^3+(-15)^3+(-13)^3\).
(i) \(-12+7+5=0\Rightarrow\) sum \(=3(-12)(7)(5)=-1260\).
(ii) \(28-15-13=0\Rightarrow\) sum \(=3(28)(-15)(-13)=16,380\).
(ii) \(28-15-13=0\Rightarrow\) sum \(=3(28)(-15)(-13)=16,380\).
15) Possible length & breadth (factor the areas): (i) \(25a^2-35a+12\) (ii) \(35y^2+13y-12\)
(i) \((5a-3)(5a-4)\).
(ii) \((7y-3)(5y+4)\).
(ii) \((7y-3)(5y+4)\).
16) Possible dimensions of cuboids:
(i) \(3x^2-12x=3x(x-4)\).
(ii) \(12ky^2+8ky-20k=4k(3y^2+2y-5)=4k(3y+5)(y-1)\).
(ii) \(12ky^2+8ky-20k=4k(3y^2+2y-5)=4k(3y+5)(y-1)\).
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