2

Q1. Find the values of a and b so that (x + 1) and (x - 2) are factors of (x³ + ax² + 2x + b).

Solution

If (x + 1) is a factor of x³ + ax² + 2x + b then (-1)³ + a(-1)² + 2(-1) + b = 0 -1 + a - 2 + b = 0 a + b = 3                   ...(1)

If (x - 2) is a factor of x³ + ax² + 2x+ b then (2)³ + a(2)² + 2(2) + b = 0 4a + b = -12 ...(2) Subtracting (1) from (2), we get 3a = -15 Or, a = -5 Using in (1), we get -5 + b = 3 Or, b = 8 Hence, a = -5, b = 8.

Q2. Without actually calculating the cubes, find the value of 75³ - 25³ - 50³.

Solution

Let x = 75, y = -25, z = -50 x + y + z = 75 - 25 - 50 = 0 We know, if x + y + z = 0 then x³ + y³ + z³ = 3xyz 75³ - 25³ - 50³ = 3(75) (-25) (-50) = 281250

Q3. Using factor theorem, show that (x + 1) is a factor of x¹⁹ + 1.

Solution

Let f(x) = x¹⁹ + 1 (x + 1) is a factor if f(-1) = 0 Now, f(-1) = (-1)¹⁹ + 1 = -1 + 1 = 0 Hence (x + 1) is a factor of x¹⁹ + 1.

Q4. Factorize : (x - y)³ + (y - z)³ + (z - x)³

Solution

Let a = x - y, b = y - z, c = z - x Here, a + b + c = x - y + y - z + z - x = 0 Now if a + b + c = 0 then x³ + y³ + z³ = 3xyz Hence, (x - y)³ + (y - z)³ + (z - x)³ = 3(x - y) (y - z) (z - x).

Q5. Factorise the polynomial x⁴ + 5x² - 6

Solution

x⁴ + 5x² - 6 Putting x² = y , the polynomial can be written as y² + 5y - 6 Factorizing it further, here a = 1, c = -6 ,    ac = -6 b = 5 = 6 - 1 y² + 5y - 6 = y² + 6y - y - 6 = y(y + 6) - 1(y + 6) = (y - 1)( y + 6) = ( x² - 1) (x² + 6)           [ putting back x² = y ] = ( x - 1) (x + 1) (x² + 6)

Q6. What are the possible expressions for the dimensions of the cuboid whose volume is given below? Volume = 12ky² + 8ky - 20k.

Solution

Volume = 12ky² + 8ky - 20k = 4k (3y² + 2y - 5) = 4k (3y² + 5y -3y - 5) = 4k (y (3y+5) - 1(3y+5)) = 4k (3y +5) (y-1) Dimensions are l = 4k, b = 3y+5, h = y-1

Q7. Factorize:(a) 4a² -9b² -2a -3b (b) a² + b² -2(ab -ac + bc)

Solution

(a) 4a²-9b²-2a-3b =[(2a)²-(3b)²]-(2a+3b) =(2a-3b) (2a+3b)-(2a+3b) =(2a+3b) [2a-3b-1] =(2a+3b) (2a-3b-1) (b) a²+b²-2(ab-ac+bc) =a²+b²-2ab+2ac-2bc =(a-b)²+2c(a-b) =(a-b)(a-b)+2c(a-b) =(a-b)(a-b+2c)

Q8. Factorise: x⁶ - 729.

Solution

x⁶ - 729 = (x²)³ - (9)³ [(x²) - (9)] [(x²)² + (x²) (9) + (9)²] = (x² - 9) (x⁴ + 9x² + 81) = (x - 3)(x + 3) (x⁴ + 9x² + 81)

Q9. If a² + b² + c² = 30 and a + b + c = 10 then find the value of ab + bc + ca.

Solution

We have (a + b + c)² = a² + +b² + c² + 2ab + 2bc + 2ca (a + b+ c)² = a² + b² + c² + 2(ab + bc + c + ca) (10)² = 30 + 2(ab + bc + ca) 100 - 30 = 2(ab + bc + ca) 70 = 2(ab + bc + ca) 35 = ab + bc + ca

Q10. Factorize : 4x⁴ + 7x² - 2

Solution

Let x² = y then equation is 4y² + 7y - 2 = 4y² + 7y - 2 = 4y² + 8y - y - 2 = 4y(y + 2) - (y + 2) = (y + 2)(4y - 1) = (x² + 2) (4x² - 1) = (x² + 2) ((2x)² - 1) = (x² + 2) (2x + 1) (2x - 1)

Q11. If x and y be two positive real number such that 8x³ + 27y³ = 730 and 2x²y + 3xy² = 15 then show that 2x + 3y = 10.

Solution

8x³ + 27y³ = 730 2x²y + 3xy² = 15 (2x + 3y)³ = (2x)³ + (3y)³ + 3 (2x) (3y) (2x + 3y) = 8x³ + 27y³ + 18xy (2x + 3y) (2x + 3y)³ = 8x³ + 27y³ + 18 (2x²y + 3xy²) (2x + 3y)³ = 730 + 18 (15) = 730 + 270 (2x + 3y)³ = 1000 2x + 3y = 10

Q12. The volume of a cube is given by the polynomial p(x) = 8x³ - 36x² + 54x - 27. Find the possible expression for the sides of cube.

Solution

8x³ - 36x² + 54x - 27 = (2x)³ + 3(2x)²(-3) + 3(2x)(-3)² + (-3)³ = (2x - 3)³ Since volume of a cube is give by (side)³. Therefore the expression for side is: 2x - 3.

Q13. Check whether the polynomial t + 1 is a factor of 4t³ + 4t² - t - 1.

Solution

p(x) = 4t³ + 4t² - t - 1 p(-1) = 4(-1)³ + 4(-1)² - (-1) - 1 = -4 + 4 + 1 - 1 = 0 Thus, (t + 1) is a factor of p(x).

Q14. Factorise: (x² - 2x)² - 11 (x² - 2x) + 24

Solution

Put x² - 2x = y So, (x² - 2x)² - 11 (x² - 2x) + 24 = y² - 11y + 24 = y² - 8y - 3y + 24 = y(y - 8) - 3(y - 8) = (y - 8) (y - 3) = (x² - 2x - 8)(x² - 2x - 3) = (x² - 4x + 2x - 8)(x² - 3x + x - 3) = (x + 2) (x - 4) (x - 3) (x + 1)

Q15. Factorize : 8 - 27x³ - 36x + 54x²

Solution

8 - 27x³ - 36x + 54x² = (2)³ + (-3x)³ + 3(2)²(-3x) + 3(2)( - 3x)² = (2 - 3x)³ = (2 - 3x) (2 - 3x) (2 - 3x)

Q16. Factorize : x³ - 8y³ - 27z³ - 18xyz.

Solution

x³ - 8y³ - 27z³ - 18xyz = (x)³ + (-2y)³ + (-3z)³ - 3(x)(-2y) (-3z) Using the identity: a³ + b³ + c³ - 3abc - (a+ b + c) (a² + b² + c² - ab - bc - ca) = (x - 2y - 3z) (x² + 4y² + 9z² - 2xy + 6yz - 3xz)

Q17. Using suitable identity evaluate (-28)³ + (9)³ + (19)³

Solution

x³ + y³ + z³ = 3xyz if x + y + z = 0 Let x = -28 y = 9 z = 19 x + y + z = (-28) + (9) + (19) = 0 (-28)³ + (9)³ + (19)³ = 3(-28) (9) (19) = -14364

Q18. Find the value of a if (x + a) is a factor of x⁴ -a²x² + 3x -a.

Solution

Let f(x) = x⁴-a²x²+3x-a (x+a) is a factor of f (x) f (-a) = 0 (-a)⁴-a²(-a)²+3(-a)-a = 0 a⁴-a⁴-3a-a = 0 -4a = 0 a = 0

Q19. Factorize : 3x² - x - 4

Solution

3x² - x - 4 = 3x² + 3x - 4x - 4 = 3x (x + 1) -4 (x + 1) = (x + 1) (3x - 4)

Q20. Factorize: a¹²y⁴ - a⁴y¹².

Solution

a¹²y⁴ -a⁴y¹² a⁴y⁴ (a⁸ - y⁸) (Taking common terms) a⁴y⁴ [(a⁴)² - (y⁴)²] a⁴y⁴ [(a⁴ + y⁴) (a⁴ -y⁴)] a⁴y⁴[(a⁴ + y⁴) (a² + y²) (a² - y²)] a⁴y⁴ [(a⁴ + y⁴) (a² + y²) (a + y) (a - y)]