answer by these questions plz
Answers
Q - 51 : Find value of p and q if (a² - 4) is a factor of pa⁴ + 2a³ - 3a² + qa - 4
Solution :
Given,
a² - 4 = 0
a² = 4
a = √4
a = ±2
pa⁴ + 2a³ - 3a² + qa - 4
i)Substitute 2 in the place of a.
p(2)⁴ + 2(2)³ - 3(2)² + q(2) - 4 = 0
p(16) + 2(8) - 3(4) + 2q - 4 = 0
16p + 16 - 12 + 2q - 4 = 0
16p + 2q + 16 - 16 = 0
16p + 2q = 0 ----(1)
ii) Substitute -2 in the place of a
p(-2)⁴ + (-2)³ - 3(-2)² + q(-2) - 4 = 0
16p + (-8) - 3(4) - 2q - 4 = 0
16p - 8 - 12 - 2q - 4 = 0
16p - 24 - 2q = 0
16p - 2q = 24 ---------(2)
Solve 1 & 2
16p + 2q = 0
16p - 2q = 24
-----------------
32p = 24
p = 24/32
p = 3/4
Substitute p in eq - (1)
16p + 2q = 0
16(3/4) + 2q = 0
4 + 2q = 0
2q = -4
q = -4/2
q = -2
Therefore, the values of p and q are 3/4 and -2 respectively.
Q - 52 : Factorise 2x³ - 3x² - 17x + 30
Solution :
2x³ - 3x² - 17x + 30
splitting the terms
2x³ - 4x² + x² - 2x - 15x + 30
2x²(x - 2) + x(x - 2) - 15(x - 2)
2x² + x - 15 ; x - 2
2x² + 6x - 5x - 15
2x(x + 3) - 5(x + 3)
(2x - 5) ; (x + 3) ; (x - 2)
Q - 53 : Multiply x² + 4y² + z² + 2xy + xz - 2yz by (-z + x - 2y)
Solution : Refer the attachement !
Q - 54 :
(i) p(2) = x² - 5x + 4
p(2) = (2)² - 5(2) + 4
p(2) = 4 - 10 + 4
p(2) = 8 - 10
p(2) = -2
ii) p(3) = x² - 5x + 4
p(3) = (3)² - 5(3) + 4
p(3) = 9 - 15 + 4
p(3) = 13 - 15
p(3) = -2
No, p(3) is not a factor.
iii) Factorise x² - 5x + 4
x² + x + 4x + 4
x(x + 1) + 4(x + 1)
(x + 4 ) ; (x + 1)
