### Algebra Class 9th Mathematics Term 1 Tamilnadu Board Solution

##### Question 1.Determine whether (x + 1) is a factor of the following polynomials:6x4 + 7x3 – 5x – 4Answer:Let f(x) = 6x4 + 7x3 – 5x – 4By factor theorem,x + 1 = 0 ;x = – 1If f(– 1) = 0 then (x + 1) is a factor of f(x)∴f(– 1) = 6(– 1)4 + 7(– 1)3 – 5(– 1) – 4= 6 – 7 + 5 – 4 = 11 – 11 = 0∴(x + 1) is a factor of f(x) = 6x4 + 7x3 – 5x – 4Question 2.Determine whether (x + 1) is a factor of the following polynomials:2x4 + 9x3 + 2x2 + 10x + 15Answer:Let f(x) = 2x4 + 9x3 + 2x2 + 10x + 15By factor theorem,x + 1 = 0 ;x = – 1If f(– 1) = 0 then (x + 1) is a factor of f(x)∴f(– 1) = 2(– 1)4 + 9(– 1)3 + 2(– 1)2 + 10(– 1) + 15= 2 – 9 + 2 – 10 + 15 = 19 – 19 = 0∴(x + 1) is a factor of f(x) = 2x4 + 9x3 + 2x2 + 10x + 15Question 3.Determine whether (x + 1) is a factor of the following polynomials:3x3 + 8x2 + 6x – 5Answer:Let f(x) = 3x3 + 8x2 + 6x – 5By factor theorem,x + 1 = 0 ;x = – 1If f(– 1) = 0 then (x + 1) is a factor of f(x)∴f(– 1) = 3(– 1)3 + 8(– 1)2 – 6(– 1) – 5= – 3 + 8 + 6 – 5 = 6(not equal to 0)∴(x + 1) is not a factor of f(x) = 3x3 + 8x2 + 6x – 5Question 4.Determine whether (x + 1) is a factor of the following polynomials:x3 – 14x2 + 3x + 12Answer:Let f(x) = x3 – 14x2 + 3x + 12By factor theorem,x + 1 = 0 ;x = – 1If f(– 1) = 0 then (x + 1) is a factor of f(x)∴f(– 1) = (– 1)3 – 14(– 1)2 + 3(– 1) + 12= – 1 – 14 – 3 + 12 = – 6(not equal to 0)∴(x + 1) is not a factor of f(x) = x3 – 14x2 + 3x + 12Question 5.Determine whether (x + 4) is a factor of x3 + 3x2 – 5x + 36.Answer:Let f(x) = x3 + 3x2 – 5x + 36.By factor theorem,x + 4 = 0: x = – 4If f(– 4) = 0, then (x + 4) is a factor∴f(– 4) = (– 4)3 + 3(– 4)2 – 5(– 4) + 36= – 64 + 48 + 20 + 36= – 64 + 104 = 40∴f(– 4) is not equal to 0So, (x + 4) is not a factor of f(x).Question 6.Using factor theorem show that (x – 1) is a factor of 4x3 – 6x2 + 9x – 7.Answer:f(x) = 4x3 – 6x2 + 9x – 7By factor theorem,(x – 1) = 0 ; x = 1Since, (x – 1) is a factor of f(x)Therefore, f(1) = 0f(1) = 4(1)3 – 6(1)2 + 9(1) – 7 = 4 – 6 + 9 – 7 = 13 – 13 = 0∴(x – 1) is a factor of f(x)Question 7.Determine whether (2x + 1) is a factor of 4x3 + 4x2 – x – 1.Answer:Let f(x) = 4x3 + 4x2 – x – 1By factor Theorem,2x + 1 = 0 ; x = – 1/2∴f(– 1/2) = 4(– 1/2)3 + 4(– 1/2)2 – (– 1/2) – 1= 4(– 1/8) + 4(1/4) + (1/2) – 1= (– 1/2) + 1 + (1/2) – 1 = 0∴f(– 1/2) = 0So, (2x + 1) is a factor of f(x).Question 8.Determine the value of p if (x + 3) is a factor of x3 – 3x2 – px + 24.Answer:Let f(x) = x3 + 3x2 – px + 24.By factor theorem,x + 3 = 0;x = – 3∴(x + 3) is a factor of f(x)So, f(– 3) = 0.f(– 3) = (– 3)3 – 3(– 3)2 – p(– 3) + 24 = 0= >27 – 27 + 3p + 24 = 0= > – 59 + 24 + 3p = 0∴3p – 30 = 0⇒ p = 30/3⇒p = 10
###### Exercise 3.6

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