Question

pls help me doing 12-15th question​

Answer 1

As a factor x² - 1 must be equal to 0.

x² - 1 = 0

x = ± 1

p(x) = x⁴ + ax³ + 2x² - 3x + b

p(1) = 0

1⁴ + a*1³ + 2*1² - 3*1 + b = 0

1 + a + 2 - 3 + b = 0

a + b = 0 _(i)

p(-1) = 0

-1⁴ + a*(-1)³ + 2*(-1)² - 3*(-1) + b = 0

1 - a + 2 + 3 + b = 0

b - a = -6 _(ii)

From (i) & (ii) , we get

• a = 3
• b = -3

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z - 3 = 0

z = 3

p(z) = az³ + 4z² + 3z - 4

p(3) = a*3³ + 4*3² + 3*3 - 4

p(3) = 27a + 36 + 9 - 4

p(3) = 27a + 41

f(z) = z³ - 4z + a

f(3) = 3³ - 4*3 + a = 15 + a

Since remainder are equal,

27a + 41 = 15 + a

26a = -26

a = -1

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

9² = a² + b² + c² + 2*26

a² + b² + c² = 81 - 52 = 29

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a³ + b³ + c³ - 3abc = ( a + b + c)(a² + b² + c² - ab - bc - ca).

8a³ - b³ + 64c³ + 24abc

(2a)³ + (-b)³ + (4c)³ - 3*(2a)*(-b)*(4c)

(2a - b + 4c)(4a² + b² + 16c² + 2ab + 4bc - 8ac)

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Answer 2

12)

$given \: {x}^{2} - 1 \: is \: a \: factor \:$

so

x=+-1

by putting x = 1 we get

p(1) = (1)^4+a(1)^3+2(1)^2-3(1)+b=0

= a+b=0 eq=1

by putting x= -1 we get

p(-1)= (-1)^4+a(-1)^3+2(-1)^2-3(-1)+b=0

-a+b=-6. eq=2

from equation 1 and 2 we get

a=3

b=(-3)

13)

given z-3=0

z=3

p(3)=a(3)^3+4(3)^2+3(3)-4

p(3)=27a+41

f(z)=z^3-4z+a

f(3)= (3)^3-4(3)+a

f(3)=15+a

as the reminder is same

26a=-26

a=(-1)

14)

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)

9^2=a^2+b^2+c^2+2(26)

a^2+b^2+c^2=81-52=29

15)

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ = {8a}^{3} - {b}^{3} + {4c}^{3} - 3(2a)( - b)(4c) \\ = (2a - b + 4c)( {4a}^{2} + {b}^{2} + {16c}^{2} + 2ab + 4bc + 8ac)$

hope it helps you ✌️✅✅✅