AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.4 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.4

### 10th Class Maths 3rd Lesson Polynomials Ex 3.4 Textbook Questions and Answers

Question 1.

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

i) p(x) = x^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2

ii) p(x) = x^{4} – 3×2 + 4x + 5, g(x) = x^{2} + 1 – x

iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

Answer:

i) Given polynomials are

p(x) = x^{3} – 3x^{2} + 5x – 3 and

g(x) = x^{2} – 2

Here, dividend and divisor are both in standard forms.

So, we have

∴ The quotient is x – 3 and the remainder is 7x – 9.

ii) Given polynomials are

p{x) = x^{4} – 3x^{2} + 4x + 5 and

g(x) = x^{2} + 1 – x

Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as x^{2} – x + 1.

We have,

∴ The quotient is x^{2} + x – 3 and the remainder is +8.

iii) Given polynomials are

p(x) = x^{4} – 5x + 6 and

g(x) = 2 – x^{2}

Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as -x^{2} + 2.

So, we have

∴ The quotient is -x^{2} – 2 and the remainder is -5x + 10.

Question 2.

Check in which case 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, x^{5} – 4x^{3} + x^{2} + 3x + 1

Answer:

i) Given first polynomial is t^{2} – 3.

Second polynomial is

2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

Let us divide 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 by t^{2} – 3, we get

Since the remainder is 0, therefore, t^{2} – 3 is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

ii) Given first polynomial is x^{2} + 3x + 1

Second polynomial is 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Let us divide 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 by x^{2} + 3x + 1, we get

Since the remainder is 0, therefore x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

iii) Given first polynomial = x^{3} – 3x + 1

Second polynomial = x^{5} – 4x^{3} + x^{2} + 3x + 1

Let us divide x^{5} – 4x^{3} + x^{2} + 3x + 1 by x^{3} – 3x + 1, we get

Here, remainder is 2(≠ 0).

Therefore, x^{3} – 3x + 1 is not a factor of x^{5} – 4x^{3} + x^{2} + 3x + 1.

Question 3.

Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).

Answer:

Let the other two zeroes are α and β.

Now compare the given polynomial 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 with the standard form ax^{4} + bx^{3} + cx^{2} + dx + e we get a = 3, b = 6, c = -2, d = -10, e = -5

–\(\frac{5}{3}\)αβ = \(\frac{-5}{3}\) ⇒ αβ = 1

now (α – β)^{2} = (α + β)^{2} – 4αβ

= (-2)^{2} – 4(1)

= 4 – 4 = 0

α – β = 0 …. (2)

Now solving (1) and (2) we get

⇒ α = -1, β = -1

Then the remaining the zeroes are -1 and -1.

Hence all zeroes of it = –\(\sqrt{\frac{5}{3}}\), \(\sqrt{\frac{5}{3}}\), -1, -1.

Question 4.

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).

Answer:

Given, p(x) = x^{3} – 3x^{2} + x + 2

q(x) = x – 2 and

r(x) = -2x + 4

By division algorithm, we know that Dividend = Divisor × Quotient + Remainder

p(x) = q(x) × g(x) + r(x)

Therefore, x^{3} – 3x^{2} + x + 2

= (x – 2) × g(x) + (- 2x + 4)

⇒ x^{3} – 3x^{2} + x + 2 + 2x – 4 = (x – 2) × g(x)

g(x) = \(\frac{x^{3}-3 x^{2}+3 x-2}{x-2}\)

On dividing x^{3} – 3x^{2} + x + 2, by x – 2, we get

First term of g(x) = \(\frac{\mathrm{x}^{3}}{\mathrm{x}}\) = x^{2}

Second term of g(x) = \(\frac{-x^{2}}{x}\) = -x

Third term of g(x) = \(\frac{x}{x}\) = 1

Hence, g(x) = x^{2} – x + 1.

Question 5.

Give examples of polynomials 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

Answer:

Let q(x) = 3x^{2} + 2x + 6, degree of q(x) = 2

p(x) = 12x^{2} + 8x + 24, degree of p(x) = 2

Given degree p(x) = degree q(x)

i) Using division algorithm,

We gave, p(x) = q(x) × g(x) + r(x)

On dividing 12x^{2} + 8x + 24 by 3x^{2} + 2x + 6, we get

Since, the remainder is zero, therefore 3x^{2} + 2x + 6 is a factor of 12x^{2} + 8x + 24.

∴ g(x) = 4 and r(x)= 0

ii) Let p(x) = x^{5} + 2x^{4} + 3x^{3} + 5x^{2} + 2

q(x) = x^{2} + x + 1, degree q(x) = 2

Given degree q(x) = degree r(x)

On dividing x^{5} + 2x^{4} + 3x^{3} + 5x^{2} + 2 by x^{2} + x + 1, we get

Here, g(x) = x^{3} + x^{2} + x + 1 and r(x) = 2x^{2} – 2x + 1

degree of r(x) = 2.

∴ deg g(x) = deg r(x).

iii) Let p(x) = 2x^{4} + 8x^{3} + 6x^{2} + 4x + 12, r(x) = 2

Here, degree r(x) = 0

On dividing 2x^{4} + 8x^{3} + 6x^{2} + 4x + 12 by 2, we get

Here, g(x) = x^{4} + 4x^{3} + 3x^{2} + 2x + 1 and r(x) = 10

so degree of r(x) = 0