Chapter 2: Polynomials
Exercise 2.1
Q1
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) \( 4x^{2}-3x+7 \)
(ii) \( y^{2}+\sqrt{2} \)
(iii) \( 3\sqrt{t}+t\sqrt{2} \)
(iv) \( y+\frac{2}{y} \)
(v) \( x^{10}+y^{3}+t^{50} \)
(i) \( 4x^{2}-3x+7 \)
(ii) \( y^{2}+\sqrt{2} \)
(iii) \( 3\sqrt{t}+t\sqrt{2} \)
(iv) \( y+\frac{2}{y} \)
(v) \( x^{10}+y^{3}+t^{50} \)
(i) Yes; this expression is a polynomial in one variable \( x \).
(ii) Yes; this expression is a polynomial in one variable \( y \).
(iii) No; the exponent of variable \( t \) in term \( 3\sqrt{t} \) is \( \frac{1}{2} \), which is not a whole number. Therefore, this expression is not a polynomial.
(iv) No; the exponent of variable \( y \) in term \( 2/y \) is -1, which is not a whole number. Therefore, this expression is not a polynomial.
(v) [Note: The source text does not explicitly provide the answer for this part, but conventionally, this is not a polynomial in one variable as it contains \(x, y, t\).]
Q2
Write the coefficients of \( x^{2} \) in each of the following:
(i) \( 2+x^{2}+x \)
(ii) \( 2-x^{2}+x^{3} \)
(iii) \( \frac{\pi}{2}x^{2}+x \)
(iv) \( \sqrt{2}x-1 \)
(i) \( 2+x^{2}+x \)
(ii) \( 2-x^{2}+x^{3} \)
(iii) \( \frac{\pi}{2}x^{2}+x \)
(iv) \( \sqrt{2}x-1 \)
(i) The coefficient of \( x^{2} \) is 1.
(ii) The coefficient of \( x^{2} \) is -1.
(iii) The coefficient of \( x^{2} \) is \( \frac{\pi}{2} \).
(iv) This can be written as \( 0.x^{2}+\sqrt{2}x-1 \). The coefficient of \( x^{2} \) is 0.
Q3
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
A binomial has two terms. Therefore, a binomial of degree 35 can be written as \( x^{35}+x^{34} \).
A monomial has only one term. Therefore, a monomial of degree 100 can be written as \( x^{100} \).
Q4
Write the degree of each of the following polynomials:
(i) \( 5x^{3}+4x^{2}+7x \)
(ii) \( 4-y^{2} \)
(iii) \( 5t-\sqrt{7} \)
(iv) 3
(i) \( 5x^{3}+4x^{2}+7x \)
(ii) \( 4-y^{2} \)
(iii) \( 5t-\sqrt{7} \)
(iv) 3
(i) 3; as the highest power of variable \( x \) is 3.
(ii) 2; as the highest power of variable \( y \) is 2.
(iii) 1; as the highest power of variable \( t \) is 1.
(iv) [Note: The degree of a non-zero constant polynomial is 0].
Q5
Classify the following as linear, quadratic and cubic polynomial:
(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) \( x^{2}+x \)
(ii) \( x-x^{3} \)
(iii) \( y+y^{2}+4 \)
(iv) \( 1+x \)
(v) \( 3t \)
(vi) \( r^{2} \)
(vii) \( 7x^{3} \)
(i) Quadratic; degree is 2.
(ii) Cubic; degree is 3.
(iii) Quadratic; degree is 2.
(iv) Linear; degree is 1.
(v) Linear; degree is 1.
(vi) Quadratic; degree is 2.
(vii) Cubic; degree is 3.
Exercise 2.2
Q1
Find the value of the polynomial \( 5x-4x^{2}+3 \) at:
(i) \( x=0 \)
(ii) \( x=-1 \)
(iii) \( x=2 \)
(i) \( x=0 \)
(ii) \( x=-1 \)
(iii) \( x=2 \)
Let \( p(x) = 5x-4x^{2}+3 \)
(i) \( p(0) = 5(0)-4(0)^{2}+3 = 3 \).
(ii) \( p(-1) = 5(-1)-4(-1)^{2}+3 = -5-4+3 = -6 \).
(iii) \( p(2) = 5(2)-4(2)^{2}+3 = 10-16+3 = -3 \).
Q2
Find \( p(0), p(1) \) and \( p(2) \) for each of the following polynomials:
(i) \( p(y)=y^{2}-y+1 \)
(ii) \( p(t)=2+t+2t^{2}-t^{3} \)
(iii) \( p(x)=x^{3} \)
(iv) \( p(x)=(x-1)(x+1) \)
(i) \( p(y)=y^{2}-y+1 \)
(ii) \( p(t)=2+t+2t^{2}-t^{3} \)
(iii) \( p(x)=x^{3} \)
(iv) \( p(x)=(x-1)(x+1) \)
(i) \( p(y)=y^{2}-y+1 \)
- \( p(0) = 0-0+1 = 1 \).
- \( p(1) = 1-1+1 = 1 \).
- \( p(2) = 4-2+1 = 3 \).
(ii) \( p(t)=2+t+2t^{2}-t^{3} \)
- \( p(0) = 2+0+0-0 = 2 \).
- \( p(1) = 2+1+2-1 = 4 \).
- \( p(2) = 2+2+8-8 = 4 \).
(iii) \( p(x)=x^{3} \)
- \( p(0) = 0 \).
- \( p(1) = 1 \).
- \( p(2) = 8 \).
(iv) \( p(x)=(x-1)(x+1) \)
- \( p(0) = (-1)(1) = -1 \).
- \( p(1) = (0)(2) = 0 \).
- \( p(2) = (1)(3) = 3 \).
Q3
Verify whether the following are zeroes of the polynomial, indicated against them:
(i) \( p(x)=3x+1, x=-\frac{1}{3} \)
(ii) \( p(x)=5x-\pi, x=\frac{4}{5} \)
(iii) \( p(x)=x^{2}-1, x=1,-1 \)
(iv) \( p(x)=(x+1)(x-2), x=-1,2 \)
(v) \( p(x)=x^{2}, x=0 \)
(vi) \( p(x)=lx+m, x=-\frac{m}{l} \)
(vii) \( p(x)=3x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \)
(viii) \( p(x)=2x+1, x=\frac{1}{2} \)
(i) \( p(x)=3x+1, x=-\frac{1}{3} \)
(ii) \( p(x)=5x-\pi, x=\frac{4}{5} \)
(iii) \( p(x)=x^{2}-1, x=1,-1 \)
(iv) \( p(x)=(x+1)(x-2), x=-1,2 \)
(v) \( p(x)=x^{2}, x=0 \)
(vi) \( p(x)=lx+m, x=-\frac{m}{l} \)
(vii) \( p(x)=3x^{2}-1, x=-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}} \)
(viii) \( p(x)=2x+1, x=\frac{1}{2} \)
(i) Yes; \( p(-\frac{1}{3}) = 3(-\frac{1}{3})+1 = -1+1 = 0 \).
(ii) No; \( p(\frac{4}{5}) = 5(\frac{4}{5})-\pi = 4-\pi \neq 0 \).
(iii) Yes; \( p(1)=0 \) and \( p(-1)=0 \).
(iv) Yes; \( p(-1)=0 \) and \( p(2)=0 \).
(v) Yes; \( p(0)=0 \).
(vi) Yes; \( p(-\frac{m}{l}) = l(-\frac{m}{l})+m = -m+m = 0 \).
(vii) Mixed; \( p(-\frac{1}{\sqrt{3}}) = 3(\frac{1}{3})-1 = 0 \), so it is a zero. But \( p(\frac{2}{\sqrt{3}}) = 3(\frac{4}{3})-1 = 3 \neq 0 \), so it is not a zero.
(viii) No; \( p(\frac{1}{2}) = 2(\frac{1}{2})+1 = 2 \neq 0 \).
Q4
Find the zero of the polynomial in each of the following cases:
(i) \( p(x)=x+5 \)
(ii) \( p(x)=x-5 \)
(iii) \( p(x)=2x+5 \)
(iv) \( p(x)=3x-2 \)
(v) \( p(x)=3x \)
(vi) \( p(x)=ax, a\neq0 \)
(vii) \( p(x)=cx+d, c\neq0 \)
(i) \( p(x)=x+5 \)
(ii) \( p(x)=x-5 \)
(iii) \( p(x)=2x+5 \)
(iv) \( p(x)=3x-2 \)
(v) \( p(x)=3x \)
(vi) \( p(x)=ax, a\neq0 \)
(vii) \( p(x)=cx+d, c\neq0 \)
To find the zero, set \( p(x)=0 \).
(i) \( x+5=0 \Rightarrow x=-5 \).
(ii) \( x-5=0 \Rightarrow x=5 \).
(iii) \( 2x+5=0 \Rightarrow x=-5/2 \).
(iv) \( 3x-2=0 \Rightarrow x=2/3 \).
(v) \( 3x=0 \Rightarrow x=0 \).
(vi) \( ax=0 \Rightarrow x=0 \).
(vii) \( cx+d=0 \Rightarrow x=-d/c \).
Exercise 2.3
Q1
Determine which of the following polynomials has \( (x+1) \) a factor:
(i) \( x^{3}+x^{2}+x+1 \)
(ii) \( x^{4}+x^{3}+x^{2}+x+1 \)
(iii) \( x^{4}+3x^{3}+3x^{2}+x+1 \)
(iv) \( x^{3}-x^{2}-(2+\sqrt{2})x+\sqrt{2} \)
(i) \( x^{3}+x^{2}+x+1 \)
(ii) \( x^{4}+x^{3}+x^{2}+x+1 \)
(iii) \( x^{4}+3x^{3}+3x^{2}+x+1 \)
(iv) \( x^{3}-x^{2}-(2+\sqrt{2})x+\sqrt{2} \)
By Factor Theorem, \( (x+1) \) is a factor if \( p(-1) = 0 \).
(i) \( p(-1) = (-1)^3+(-1)^2+(-1)+1 = -1+1-1+1 = 0 \). Yes..
(ii) \( p(-1) = 1-1+1-1+1 = 1 \neq 0 \). No..
(iii) \( p(-1) = 1-3+3-1+1 = 1 \neq 0 \). No..
(iv) \( p(-1) = -1-1+2+\sqrt{2}+\sqrt{2} = 2\sqrt{2} \neq 0 \). No..
Q2
Use the Factor Theorem to determine whether \( g(x) \) is a factor of \( p(x) \):
(i) \( p(x)=2x^{3}+x^{2}-2x-1, g(x)=x+1 \)
(ii) \( p(x)=x^{3}+3x^{2}+3x+1, g(x)=x+2 \)
(iii) \( p(x)=x^{3}-4x^{2}+x+6, g(x)=x-3 \)
(i) \( p(x)=2x^{3}+x^{2}-2x-1, g(x)=x+1 \)
(ii) \( p(x)=x^{3}+3x^{2}+3x+1, g(x)=x+2 \)
(iii) \( p(x)=x^{3}-4x^{2}+x+6, g(x)=x-3 \)
(i) Check \( p(-1) \): \( 2(-1)^3+(-1)^2-2(-1)-1 = -2+1+2-1 = 0 \). Yes..
(ii) Check \( p(-2) \): \( -8+12-6+1 = -1 \neq 0 \). No..
(iii) Check \( p(3) \): \( 27-36+3+6 = 0 \). Yes..
Q3
Find the value of \( k \), if \( x-1 \) is a factor of \( p(x) \):
(i) \( p(x)=x^{2}+x+k \)
(ii) \( p(x)=2x^{2}+kx+\sqrt{2} \)
(iii) \( p(x)=kx^{2}-\sqrt{2}x+10 \)
(iv) \( p(x)=kx^{2}-3x+k \)
(i) \( p(x)=x^{2}+x+k \)
(ii) \( p(x)=2x^{2}+kx+\sqrt{2} \)
(iii) \( p(x)=kx^{2}-\sqrt{2}x+10 \)
(iv) \( p(x)=kx^{2}-3x+k \)
Since \( x-1 \) is a factor, \( p(1)=0 \).
(i) \( 1^2+1+k=0 \Rightarrow k=-2 \).
(ii) \( 2(1)^2+k(1)+\sqrt{2}=0 \Rightarrow k=-(2+\sqrt{2}) \).
(iii) \( k(1)^2-\sqrt{2}(1)+1=0 \Rightarrow k=\sqrt{2}-1 \).
(iv) \( k(1)^2-3(1)+k=0 \Rightarrow 2k=3 \Rightarrow k=3/2 \).
Q4
Factorise:
(i) \( 12x^{2}-7x+1 \)
(ii) \( 2x^{2}+7x+3 \)
(iii) \( 6x^{2}+5x-6 \)
(iv) \( 3x^{2}-x-4 \)
(i) \( 12x^{2}-7x+1 \)
(ii) \( 2x^{2}+7x+3 \)
(iii) \( 6x^{2}+5x-6 \)
(iv) \( 3x^{2}-x-4 \)
(i) \( 12x^{2}-7x+1 \)
Split middle term: \( 12x^2 - 4x - 3x + 1 \)
\( = 4x(3x-1)-1(3x-1) = (3x-1)(4x-1) \).
(ii) \( 2x^{2}+7x+3 \)
Split middle term: \( 2x^2 + 6x + x + 3 \)
\( = 2x(x+3)+1(x+3) = (x+3)(2x+1) \).
(iii) \( 6x^{2}+5x-6 \)
Split middle term: \( 6x^2 + 9x - 4x - 6 \)
\( = 3x(2x+3)-2(2x+3) = (2x+3)(3x-2) \).
(iv) \( 3x^{2}-x-4 \)
Split middle term: \( 3x^2 - 4x + 3x - 4 \)
\( = x(3x-4)+1(3x-4) = (3x-4)(x+1) \).
Q5
Factorise:
(i) \( x^{3}-2x^{2}-x+2 \)
(ii) \( x^{3}-3x^{2}-9x-5 \)
(iii) \( x^{3}+13x^{2}+32x+20 \)
(iv) \( 2y^{3}+y^{2}-2y-1 \)
(i) \( x^{3}-2x^{2}-x+2 \)
(ii) \( x^{3}-3x^{2}-9x-5 \)
(iii) \( x^{3}+13x^{2}+32x+20 \)
(iv) \( 2y^{3}+y^{2}-2y-1 \)
(i) Using factor theorem, \( p(2)=0 \), so \( (x-2) \) is a factor. Dividing by \( (x-2) \) gives quotient \( x^2-1 \).
Factorising quotient: \( (x-1)(x+1) \).
Result: \( (x-2)(x-1)(x+1) \).
(ii) Using factor theorem, \( p(-1)=0 \), so \( (x+1) \) is a factor. Dividing by \( (x+1) \) gives quotient \( x^2-4x-5 \).
Factorising quotient: \( (x-5)(x+1) \).
Result: \( (x-5)(x+1)(x+1) \).
(iii) Using factor theorem, \( p(-1)=0 \), so \( (x+1) \) is a factor. Dividing by \( (x+1) \) gives quotient \( x^2+12x+20 \).
Factorising quotient: \( (x+10)(x+2) \).
Result: \( (x+1)(x+2)(x+10) \).
(iv) Using factor theorem, \( p(1)=0 \), so \( (y-1) \) is a factor. Dividing by \( (y-1) \) gives quotient \( 2y^2+3y+1 \).
Factorising quotient: \( (y+1)(2y+1) \).
Result: \( (y-1)(y+1)(2y+1) \).
Exercise 2.4
Q1
Use suitable identities to find the following products:
(i) \( (x+4)(x+10) \)
(ii) \( (x+8)(x-10) \)
(iii) \( (3x+4)(3x-5) \)
(iv) \( (y^{2}+\frac{3}{2})(y^{2}-\frac{3}{2}) \)
(v) \( (3-2x)(3+2x) \)
(i) \( (x+4)(x+10) \)
(ii) \( (x+8)(x-10) \)
(iii) \( (3x+4)(3x-5) \)
(iv) \( (y^{2}+\frac{3}{2})(y^{2}-\frac{3}{2}) \)
(v) \( (3-2x)(3+2x) \)
(i) \( x^{2}+(4+10)x+40 = x^{2}+14x+40 \).
(ii) \( x^{2}+(8-10)x-80 = x^{2}-2x-80 \).
(iii) \( 9[x^2 - 1/3x - 20/9] = 9x^{2}-3x-20 \).
(iv) \( (y^{2})^{2}-(\frac{3}{2})^{2} = y^{4}-\frac{9}{4} \).
(v) \( 3^{2}-(2x)^{2} = 9-4x^{2} \).
Q2
Evaluate the following products without multiplying directly:
(i) \( 103\times107 \)
(ii) \( 95\times96 \)
(iii) \( 104\times96 \)
(i) \( 103\times107 \)
(ii) \( 95\times96 \)
(iii) \( 104\times96 \)
(i) \( (100+3)(100+7) = 10000+1000+21 = 11021 \).
(ii) \( (100-5)(100-4) = 10000-900+20 = 9120 \).
(iii) \( (100+4)(100-4) = 10000-16 = 9984 \).
Q3
Factorise the following using appropriate identities:
(i) \( 9x^{2}+6xy+y^{2} \)
(ii) \( 4y^{2}-4y+1 \)
(iii) \( x^{2}-\frac{y^{2}}{100} \)
(i) \( 9x^{2}+6xy+y^{2} \)
(ii) \( 4y^{2}-4y+1 \)
(iii) \( x^{2}-\frac{y^{2}}{100} \)
(i) \( (3x)^{2}+2(3x)(y)+(y)^{2} = (3x+y)^{2} \).
(ii) \( (2y)^{2}-2(2y)(1)+(1)^{2} = (2y-1)^{2} \).
(iii) \( x^{2}-(\frac{y}{10})^{2} = (x+\frac{y}{10})(x-\frac{y}{10}) \).
Q4
Expand each of the following, using suitable identities:
(i) \( (x+2y+4z)^{2} \)
(ii) \( (2x-y+z)^{2} \)
(iii) \( (-2x+3y+2z)^{2} \)
(iv) \( (3a-7b-c)^{2} \)
(v) \( (-2x+5y-3z)^{2} \)
(vi) \( [\frac{1}{4}a-\frac{1}{2}b+1]^{2} \)
(i) \( (x+2y+4z)^{2} \)
(ii) \( (2x-y+z)^{2} \)
(iii) \( (-2x+3y+2z)^{2} \)
(iv) \( (3a-7b-c)^{2} \)
(v) \( (-2x+5y-3z)^{2} \)
(vi) \( [\frac{1}{4}a-\frac{1}{2}b+1]^{2} \)
Using \( (x+y+z)^{2} = x^2+y^2+z^2+2xy+2yz+2zx \):
(i) \( x^{2}+4y^{2}+16z^{2}+4xy+16yz+8xz \).
(ii) \( 4x^{2}+y^{2}+z^{2}-4xy-2yz+4xz \).
(iii) \( 4x^{2}+9y^{2}+4z^{2}-12xy+12yz-8xz \).
(iv) \( 9a^{2}+49b^{2}+c^{2}-42ab+14bc-6ac \).
(v) \( 4x^{2}+25y^{2}+9z^{2}-20xy