Math, asked by aniabi614, 3 months ago

Please finish these sums I want to submit tomorrow

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Answers

Answered by karmaan958
2

Step-by-step explanation:

1. Find the remainder when 2x² - 5x - 3 is divided by

x - 3

Solution : p(x) = 2x² - 5x - 3 & g(x) = x - 3

g(x) = x - 3

x - 3 = 0

x = 3

putting the value of x in p(x)

= 2x² - 5x - 3

= 2(3)² - 5(3) - 3

= 18 - 15 - 3

= 18 - 18

= 0

remainder = 0

2. Find the value of k, if (x - 1) is a factor of

p(x) = 4x² + 3x² - 4x + k

Solution : p(x) = (4x² + 3x² - 4x + k) & g(x) = (x - 1)

g(x) = (x - 1)

x - 1 = 0

x = 1

p(x) is a factor of g(x) [given]

p(x) = 0

(4x² + 3x² - 4x + k) = 0

putting the value of x in p(x)

4(1)² + 3(1)² - 4(1) + k = 0

4 + 3 - 4 + k = 0

3 + k = 0

k = -3

3. If x² + Kx + 6 = (x + 2)(x + 3)

x² + Kx + 6 = x² + 3x + 2x + 6

x² + Kx + 6 = x² + 5x + 6

k = 5

4. p(t) = t² - t + 2 , find p(1/3)

= (1/3)² - 1/3 + 2

= 1/9 - 1/3 + 2

= 1/9 - 3/9 + 18/9

= (1 - 3 + 18)/9

= 16/9

5. determine whether indicated numbers are zeroes of the polynomial f(x) = x³ - 6x² + 11x - 6 ; x = 1, 3.

Solution :

putting value of x = 1 in f(x)

x³ - 6x² + 11x - 6

= (1)³ - 6(1)² + 11(1) - 6

= 1 - 6 + 11 - 6

= 0

1 is the zeroes of the polynomial

putting value of x = 3 in f(x)

x³ - 6x² + 11x - 6

= (3)³ - 6(3)² + 11(3) - 6

= 27 - 54 + 33 - 6

= 60 - 60

= 0

3 is also the zeroes of the polynomial

6. factorise x² - (y/100)²

Identities a² - b² = (a + b)(a - b)

x² - (y/100)²

= (x + y/10)(x - y/10)

7. (0.75 × 0.75 × 0.75 + 0.25 × 0.25 × 0.25)/(0.75 × 0.75 - 0.75 × 0.25 + 0.25 × 0.25)

= {(0.75)³ + (0.25)³}/(0.75)² - 0.75 × 0.25 + (0.25)²}

[(a + b)(a² – ab + b²) = a³ + b³]

= [(0.75 + 0.25){(0.75)² - 0.75 × 0.25 + (0.25)²]/(0.75)² - 0.75 × 0.25 + (0.25)²

= (0.75 + 0.25)

= 1.00

= 1

L.H.S = R.H.S.

hence proved

8. If x + 1/x = 3, find x² + 1/x²

Solution :

Squaring both sides

(x + 1/x)² = (3)²

x² + 1/x² + 2(x)(1/x) = 9

x² + 1/x² + 2 = 9

x² + 1/x² = 9 - 2

x² + 1/x² = 7

9. If -1 is a zero of the polynomial p(x) = ax³ - x² + x + 4,

find the value of a.

ax³ - x² + x + 4 = 0

putting value of x = (-1)

a(-1)³ - (-1)² + (-1) + 4 = 0

-1a - 1 - 1 + 4 = 0

-1a + 2 = 0

-1a = -2

a = (-2)/(-1)

a = 2

10. factorise 7√2x² - 10x - 4√2

= 7√2x² - 14x + 4x - 4√2

= 7√2x(x - √2) + 4(x - √2)

= (7√2x + 4)(x - √2)

11. x² + 3√3x + 6

= x² + 2√3x + √3x + 6

= x(x + 2√3) + √3(x + 2√3)

= (x + √3)(x + 2√3)

12. expand (x - 2y - 3z)²

[(a - b - c)² = a² + b² + c² - 2ab + 2bc - 2ac]

= (x - 2y - 3z)²

= x² + 4y² + 9z² -2(x)(2y) + 2(2y)(3z) - 2(x)(3z)

= x² + 4y² + 9z² - 4xy + 12yz - 6xz

13. Simplify (x + 1)³ - (x - 1)³

(x + 1)³ - (x - 1)³

= {x³ + (1)³ + 3(x)(1)(x + 1)} - {x³ - (1)³ - 3(x)(1)(x - 1)}

= (x³ + 1 + 3x² + 3x) - (x³ - 1 - 3x² + 3x)

= x³ + 1 + 3x² + 3x - x³ + 1 + 3x² - 3x

= 2 + 6x²

14. Evaluate (99)³

(99)³

= (100 - 1)³

[(a + b)³ = a³ + b³ + 3ab(a + b)]

= [(100)³ + (1)³ + 3(100)(1)(100 + 1)]

= 1000000 + 1 + 30000 + 300

= 1030301

15. factorise a² + b² - 2(ab - ac + bc)

= a² + b² - 2(ab - ac + bc) + c² - c²

= (a - b + c)² - (c)²

= (a - b + c + c)(a - b + c - c)

= (a - b + 2c)(a - b)

16. if x + 1/x = 7 , find the value of x³ + 1/x³

(x + 1/x)³ = (7)³

x³ + 1/x³ + 3(x)(1/x)(x + 1/x) = 343

x³ + 1/x³ + 3(7) = 343

x³ + 1/x³ + 21 = 343

x³ + 1/x³ = 343 - 21

x³ + 1/x³ = 322

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