Polynomials

EXERCISE-2.1

P1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Sol. (i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

EXERCISE-2.2 [PAGE NO. 33]

P1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x² -2x -8     (ii) 4s² -4s + 1     (iii) 6x² -3-7x     (iv) 4u² + 8u     (v) t² -15     (vi) 3x² -x -4

Sol. (i) $x^2–2x–8=(x–4)(x+2)$

The value of $x^2–2x–8=0$ is zero

when x − 4 = 0 or x + 2 = 0,

i.e., when x = 4 or x = −2

Therefore, the zeroes of $x^2–2x–8$ are 4 and −2.

Sum of zeroes = $4–2=2=\frac{–\left(2\right)}{1}=\frac{–\left(\text{ Coefficient of }\mathrm{x}\right)}{\text{ Coefficient of }{\mathrm{x}}^2}$

Product of zeroes $=4\times (–2)=–8=\frac{(–8)}{1}$$=\frac{\text{ Constant term }}{\text{ Coefficient of }{\mathrm{x}}^2}$

(ii) $4s^2–4s+1=(2s–1{)}^2$

The value of 4s² − 4s + 1 is zero, when 2s − 1 = 0, i.e., $s=\frac{1}{2}$

Therefore, the zeroes of 4s² − 4s + 1 are $\frac{1}{2}$ and $\frac{1}{2}$.

Sum of zeroes = $\frac{1}{2}+\frac{1}{2}=1=\frac{–(–4)}{4}$=$\frac{–\left(\text{ Coefficient of }s\right)}{\text{ Coefficient of }{s}^2}$

Product of zeroes $=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}=\frac{\text{ Constant term }}{\text{ Coefficient of }{\mathrm{s}}^2}$

(iii) $6x^2–3–7x=6x^2–7x–3$$=(3 x+1)(2 x–3)$

The value of 6x² − 3 − 7x is zero when 3x + 1 = 0 or 2x − 3 = 0, i.e., $x=\frac{–1}{3}$ or $x=\frac{3}{2}$

Therefore, the zeroes of 6x² − 3 − 7x are $\frac{–1}{3}$and $\frac{3}{2}$.

Sum of zeroes = $\frac{–1}{3}+\frac{3}{2}=\frac{–(–7)}{6}$= $\frac{–\left(\text{ Coefficient of }x\right)}{\text{ Coefficient of }{x}^2}$

Product of zeroes = $\frac{–1}{3}\times \frac{3}{2}=\frac{–1}{2}=\frac{–3}{6}=\frac{\text{ Constant term }}{\text{ Coefficient of }{\mathrm{x}}^2}$

(iv) $4u^2+8u=4u^2+8u+0$ $=4u\left(u+2\right)$

The value of 4u² + 8u is zero

when 4u = 0 or u + 2 = 0,

i.e., u = 0 or u = −2

Therefore, the zeroes of 4u² + 8u are 0 and −2.

Sum of zeroes = $0+(–2)=–2=\frac{–\left(8\right)}{4}$=$\frac{–\left(\text{ Coefficient of }\mathrm{u}\right)}{\text{ Coefficient of }{\mathrm{u}}^2}$

Product of zeroes = $0\times (–2)=0=\frac{0}{4}=\frac{\text{ Constant term }}{\text{ Coefficient of }{\mathrm{u}}^2}$

(v) $t^2–15$ $=t^2–0t–15$ $=\left(t–\sqrt{15}\right)\left(t+\sqrt{15}\right)$

The value of t² − 15 is zero

when  $t–\sqrt{15}=0$ or $t+\sqrt{15}=0$,

i.e., when  $t=\sqrt{15}$ or $t=–\sqrt{15}$

Therefore, the zeroes of t² − 15 are $\sqrt{15}$ and $–\sqrt{15}$.

Sum of zeroes = $\sqrt{15}+(–\sqrt{15})=0=\frac{–0}{1}$ = $\frac{–\left(\text{ Coefficient of }\mathrm{t}\right)}{\text{ Coefficient of }{\mathrm{t}}^2}$

Product of zeroes  = $\sqrt{15}(–\sqrt{15})=–15=\frac{–15}{1}$ $=\frac{\text{ Constant term }}{\text{ Coefficient of }{\mathrm{t}}^2}$

(vi)   $3x^2–x–4$

$\left(3x–4\right)\left(x+1\right)$

The value of 3x² − x − 4 is zero

when 3x − 4 = 0 or x + 1 = 0,

i.e., when $x=\frac{4}{3}$ or x = −1

Therefore, the zeroes of 3x² − x − 4 are $\frac{4}{3}$ and −1.

Sum of zeroes = $\frac{4}{3}+(–1)=\frac{1}{3}=\frac{–(–1)}{3}$ = $\frac{–\left(\text{ Coefficient of }x\right)}{\text{ Coefficient of }{x}^2}$

Product of zeroes $\frac{4}{3}\times (–1)=\frac{–4}{3}=\frac{\text{ Constant term }}{\text{ Coefficient of }{x}^2}$

P2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) $\frac{1}{4}, –1$     (ii) $\sqrt{2}, \frac{1}{3}$     (iii) $0, \sqrt{5}$     (iv) $1, 1$     (v) $\frac{–1}{4}, \frac{1}{4}$     (vi) $4, 1$

Sol. (i) $\frac{1}{4},–1$

Let the polynomial be $a{x}^2+bx+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha +\beta =\frac{1}{4}=\frac{–b}{a}$

$\alpha \beta =–1=\frac{–4}{4}=\frac{c}{a}$

If a = 4, then b = -1, c = -4

Therefore, the quadratic polynomial is 4x² − x − 4.

(ii) $\sqrt{2}, \frac{1}{3}$

Let the polynomial be $a{x}^2+bx+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha +\beta =\sqrt{2}=\frac{3\sqrt{2}}{3}=\frac{–b}{a}$

$\alpha \beta =\frac{1}{3}=\frac{c}{a}$

Therefore, the quadratic polynomial is $3x^2–3\sqrt{2}x+1$.

(iii) $0, \sqrt{5}$

Let the polynomial be $a{x}^2+bx+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha +\beta =0=\frac{0}{1}=\frac{–b}{a}$

$\alpha \beta =\sqrt{5}=\frac{\sqrt{5}}{1}=\frac{c}{a}$

If a = 1, then b = 0, c = $\sqrt{5}$

Therefore, the quadratic polynomial is $x^2+\sqrt{5}$.

(iv) 1, 1

Let the polynomial be $a{x}^2+bx+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha +\beta =1=\frac{1}{1}=\frac{–b}{a}$

$\alpha \times \beta =1=\frac{1}{1}=\frac{c}{a}$

If a = 1, then b = -1, c = 1

Therefore, the quadratic polynomial is $x^2–x+1$.

(v) $\frac{–1}{4}, \frac{1}{4}$

Let the polynomial be $a{x}^2+bx+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha +\beta =\frac{–1}{4}=\frac{–b}{a}$

$\alpha \times \beta =\frac{1}{4}=\frac{c}{a}$

If a = 4, then b = 1, c = 1

Therefore, the quadratic polynomial is 4x² + x + 1.

(vi) 4, 1

Let the polynomial be $a{x}^2+bx+c$, and its zeroes be $\alpha$ and $\beta$.

$\alpha +\beta =4=\frac{4}{1}=\frac{–b}{a}$

$\alpha \times \beta =1=\frac{1}{1}=\frac{c}{a}$

If a =1, then b = -4, c = 1

Therefore, the quadratic polynomial is x² -4x + 1.

EXERCISE-2.3

P1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) $p\left(x\right)=x^3–3x^2+5x–3, q\left(x\right)=x^2–2$

(ii) $p\left(x\right)=x^4–3x^2+4x+5, q\left(x\right)=x^2+1–x$

(iii) $p\left(x\right)=x^4–5x+6, q\left(x\right)=2–x^2$

Sol. (i) $p\left(x\right)=x^3–3x^2+5x–3, q\left(x\right)=x^2–2$

Quotient = x − 3

Remainder = 7x – 9

(ii) $p\left(x\right)=x^4–3x^2+4x+5, q\left(x\right)=x^2+1–x$

Quotient = x² + x − 3

Remainder = 8

(iii) $p\left(x\right)=x^4–5x+6, q\left(x\right)=2–x^2$

Quotient = −x² − 2

Remainder = −5x +10

P2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) $t^2–3, 2t^4+3t^3–2t^2–9t–12$

(ii) $x^2+3x+1, 3x^4+5x^3–7x^2+2x+2$

(iii) $x^3–3x+1, 3x^5–4x^3+x^2+3x+1$

Sol. (i) ${\mathrm{t}}^2–3, 2{\mathrm{t}}^4+3{\mathrm{t}}^3–2{\mathrm{t}}^2–9\mathrm{t}–12$

$t^2–3=t^2+0.t–3$

Since the remainder is 0,

Hence, t² -3  is a factor of 2t⁴ + 3t³-2t² -9t -12.

(ii) $x^2+3x+1, 3x^4+5x^3–7x^2+2x+2$

Since the remainder is 0,

Hence,  $x^2+3x+1$ is a factor of $3x^4+5x^3–7x^2+2x+2$.

(iii)  $x^3–3x+1, 3x^5–4x^3+x^2+3x+1$

Since the remainder $\ne 0$,

Hence,  $x^3–3x+1$ is not a factor of $3x^5–4x^3+x^2+3x+1$.

P3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are .

Sol. P(x) = 3x⁴ + 6x³ – 2x² -10x -5

Since the two zeroes are $\sqrt{\frac{5}{3}}$ and $–\sqrt{\frac{5}{3}}$,

$\therefore \left(x–\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=\left(x^2–\frac{5}{3}\right)$ is a factor of 3x⁴ + 6x³ – 2x² -10x -5.

Therefore, we divide the given polynomial by $x^2–\frac{5}{3}$.

$3x^4+6x^3–2x^2–10x–5$ $=\left(x^2–\frac{5}{3}\right)\left(3x^2+6x+3\right)$ $=3\left(x^2–\frac{5}{3}\right)\left(x^2+2x+1\right)$

We factorize $x^2+2x+1=(x+1{)}^2$

Therefore, its zero is given by x + 1 = 0

x = −1

As it has the term ${\left(x+1\right)}^2$, therefore, there will be 2 zeroes at x = −1.

Hence, the zeroes of the given polynomial are $\sqrt{\frac{5}{3}}, –\sqrt{\frac{5}{3}}, –1$, and −1.

P4. On dividing $x^3–3x^2+x+2$ by a polynomial g(x), the quotient and remainder were x − 2 and − 2x + 4, respectively. Find g(x).

Sol. $p\left(x\right)=x^3–3x^2+x+2$ (Dividend)

g(x) = ? (Divisor)

Quotient = (x − 2)

Remainder = (− 2x + 4)

Dividend = Divisor × Quotient + Remainder

$x^3–3x^2+x+2$ $=g\left(x\right)\times (x–2)+(–2x+4)$

$x^3–3x^2+x+2+2x–4=g\left(x\right)(x–2)$

$x^3–3x^2+3x–2=g\left(x\right)(x–2)$

g(x) is the quotient when we divide $x^3–3x^2+3x–2\text{ by }(x–2)$

$\therefore g\left(x\right)=\left(x^2–x+1\right)$

P5. Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Sol. According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that

p(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x)

Degree of a polynomial is the highest power of the variable in the polynomial.

(i) deg p(x) = deg q(x)

Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant).

Let us assume the division of by 2.

Here, p(x) = $6x^2+2x+2$

g(x) = 2

$q\left(x\right)=3x^2+x+1$ and r(x) = 0

Degree of p(x) and q(x) is the same i.e., 2.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

$6x^2+2x+2=2\left(3x^2+x+1\right)$ = $6x^2+2x+2$

Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)

Let us assume the division of x³ + x by x²,

Here, p(x) = x³ + x

g(x) = x²

q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x³ + x = (x² ) × x + x

x³ + x = x³ + x

Thus, the division algorithm is satisfied.

(iii) deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of x³ + 1by x².

Here, p(x) = x³ + 1

g(x) = x²

q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0.

Checking for division algorithm,

p(x) = g(x) × q(x) + r(x)

x³ + 1 = (x² ) × x + 1

x³ + 1 = x³ + 1

Thus, the division algorithm is satisfied.

EXERCISE-2.4

P1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

(i) $2x^3+x^2–5x+2; \frac{1}{2},1,–2$

(ii) $x^3–4x^2+5x–2; 2,1,1$

Sol. (i) $p\left(x\right)=2x^3+x^2–5x+2$

Zeroes for this polynomial are $\frac{1}{2},1,–2$

$\mathrm{p}\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{2}\right)}^2–5\left(\frac{1}{2}\right)+2$ $=\frac{1}{4}+\frac{1}{4}–\frac{5}{2}+2=0$

$\mathrm{P}\left(1\right)=2\times 1^3+1^2–5\times 1+2=0$

$\mathrm{P}(–2)=2\times (–2{)}^3+(2{)}^2–5\times (–2)+2=0$

Therefore, $\frac{1}{2}$, 1, and −2 are the zeroes of the given polynomial.

Comparing the given polynomial with $a{x}^3+b{x}^2+cx+d$, we obtain 

a = 2, b = 1, c = −5, d = 2

We can take $\alpha =\frac{1}{2}, \beta =1, \gamma =–2$

$\alpha +\beta +\gamma =\frac{1}{2}+1+(–2)$ $=–\frac{1}{2}=\frac{–b}{a}$

$\alpha \beta +\beta \gamma +\alpha \gamma =\frac{1}{2}\times 1+1\times (–2)+\frac{1}{2}(–2)=\frac{–5}{2}=\frac{c}{a}$

$\alpha \beta \gamma =\frac{1}{2}\times 1\times (–2)=$$\frac{–1}{1}=\frac{–\left(2\right)}{2}=\frac{–\mathrm{d}}{\mathrm{a}}$

Therefore, the relationship between the zeroes and the coefficients is verified.

(ii) $p\left(x\right)=x^3–4x^2+5x–2$

Zeroes for this polynomials are 2, 1, 1.

p(2) = 2³ -4(2²) + 5(1) – 2 = 8 -16 + 10 -2 = 0

p(1) = 1³ -4(1²) + 5(1) – 2 = 1 -4 + 5 -2 = 0

Therefore, 2, 1, 1 are the zeroes of the given polynomial.

Comparing the given polynomial with a , we obtain a = 1, b = −4, c = 5, d = −2.

Verification of the relationship between zeroes and coefficient of the given polynomial

Sum of zeroes = 2 + 1 + 1 = 4 = $\frac{–(–4)}{1}=\frac{–b}{a}$

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1) =2 + 1 + 2 = 5 

Multiplication of zeroes = 2 × 1 × 1 = 2 $=\frac{–(–2)}{1}=\frac{–\mathrm{d}}{\mathrm{a}}$

Hence, the relationship between the zeroes and the coefficients is verified.

P2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, − 7, − 14 respectively.

Sol. Let the polynomial be $a{x}^3+b{x}^2+cx+d$ and the zeroes be $\alpha , \beta$ and $\gamma$.

It is given that

$\alpha +\beta +\gamma =\frac{2}{1}=\frac{–b}{a}$

$\alpha \beta +\beta \gamma +\alpha \gamma =\frac{–7}{1}=\frac{c}{a}$

$\alpha \beta \gamma =\frac{–14}{1}=\frac{–\mathrm{d}}{\mathrm{a}}$

If a = 1, then b = −2, c = −7, d = 14

Hence, the polynomial is $x^3–2x^2–7x+14$.

P3. If the zeroes of polynomial $x^3–3x^2+x+1$ are a − b, a + a + b, find a and b.

Sol. $p\left(x\right)=x^3–3x^2+x+1$

Zeroes are a − b, a + a + b

Comparing the given polynomial with $p{x}^3+q{x}^2+rx+t$, we obtain

p = 1, q = −3, r = 1, t = 1

Sum of zeroes = a − b + a + a + b

$\frac{–\mathrm{q}}{\mathrm{p}}=3\mathrm{a}$

$\frac{–(–3)}{1}=3\mathrm{a}$

3 = 3a

a = 1

The zeroes are 1 – b, 1, 1 + b.

Multiplication of zeroes = 1(1 – b)(1 + b)

$\frac{–t}{p}=1–b^2$

$\frac{–1}{1}=1–b^2$

$1–b^2=–1$

$1+1=b^2$

$b=\pm \sqrt{2}$

Hence, a = 1 and b = $\sqrt{2}$ or $–\sqrt{2}$.

P4. It two zeroes of the polynomial $x^4–6x^3–26x^2+138x–35$ are $2\pm \sqrt{3}$, find other zeroes.

Sol. Given that  $2+\sqrt{3}$ and $2–\sqrt{3}$are zeroes of the given polynomial.

Therefore,  $(x–2–\sqrt{3})(x–2+\sqrt{3})$ = x² + 4 ­­− 4x − 3

= x² ­− 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing  $x^4–6x^3–26x^2+138x–35\text{ by }{x}^2–4x+1$.

Clearly, $x^4–6x^3–26x^2+138x–35$ $=\left(x^2–4x+1\right)\left(x^2–2x–35\right)$

It can be observed that $\left(x^2–2x–35\right)$ is also a factor of the given polynomial.

And $\left(x^2–2x–35\right)=(x–7)(x+5)$

Therefore, the value of the polynomial is also zero when x – 7 = 0 or x + 5 = 0

or x = 7 or − 5

Hence, 7 and −5 are also zeroes of this polynomial.

P5. If the polynomial $x^4–6x^3+16x^2–25x+10$ is divided by another polynomial $x^2–2x+k$, the remainder comes out to be x + a, find k and a.

Sol. By division algorithm,

Dividend = Divisor × Quotient + Remainder

Dividend − Remainder = Divisor × Quotient

$x^4–6x^3+16x^2–25x+10–x–a=x^4–6x^3–26x+10–a$

will be perfectly divisible by $x^2–2x+k$.

Let us divide $x^4–6x^3+16x^2–26x+10–a$ by $x^2–2x+k$

It can be observed that $(–10+2k)x+\left(10–a–8k+k^2\right)$ will be 0.

Therefore, (- 10 + 2k) = 0 and $\left(10–a–8k+k^2\right)=0$

For (- 10 + 2k) = 0,

2 k =10

And thus, k = 5

For $\left(10–a–8k+k^2\right)=0$

10 − a − 8 × 5 + 25 = 0

10 − a − 40 + 25 = 0

− 5 − a = 0

Therefore, a = −5

Hence, k = 5 and a = −5