### Algebra Class 9th Mathematics Term 2 Tamilnadu Board Solution

##### Question 1.The expansion of (x + 2)(x – 1) isA. x2 – x – 2B. x2 + x + 2C. x2 + x – 2D. x2 – x + 2Answer:Using identity,(x + a)(x + b) = x2 + (a + b)x + abWe have,(x + 2)(x – 1)= x2 + (2 – 1) x + (2x – 1) = x2 + x – 2Question 2.The expansion of (x + 1)(x – 2)(x + 3) isA. x3 + 2x2 – 5x – 6B. x3 – 2x2 + 5x – 6C. x3 + 2x2 + 5x – 6D. x3 + 2x2 + 5x + 6Answer:Using identity,(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abcWe have,(x + 1)(x – 2)(x + 3) = x3 + (1 – 2 + 3)x2 + (– 2 – 6 + 3)x – 6= x3 + 2x2 – 5x – 6Question 3.(x – y)(x2 + xy + y2) is equal toA. x3 + y3B. x2 + y2C. x2 – y2D. x3 – y3Answer:(x – y)(x2 + xy + y2) is simply the expansion of x3 – y3.we can also obtain this result by multiplying themi.e. (x – y)(x2 + xy + y2) = x(x2 + xy + y2) – y(x2 + xy + y2)= x3 + x2y + xy2 – x2y – xy2 – y3= x3 – y3Question 4.Factorization of x2 + 2x – 8 isA. (x + 4)(x – 2)B. (x – 4)(x + 2)C. (x + 4)(x + 2)D. (x – 4)(x – 2)Answer:Suppose (x + p)(x + q) are two factors of x2 + 2x – 8.Then, x2 + 2x – 8 = (x + p)(x + q)= x2 + (p + q)x + pqSo, to factorize we have to find p and q, such that pq = – 8 and p + q = 2. ∴ x2 + 2x – 8 = (x + 4)(x – 2)Question 5.If one of the factors of x2 – 6x – 16 is (x + 2) then other factor isA. x + 5B. x – 5C. x + 8D. x – 8Answer:Let the other factor be (x + p)Then, x2 – 6x – 16 = (x + p)(x + 2)= x2 + (p + 2)x + 2pOn comparing the coefficients on both sides, we get,p + 2 = – 6⇒ p = – 8∴ The other factor is (x – 8).Question 6.If (2x + 1) and (x – 3) are the factors of ax2 – 5x + c, then the values of a and c are respectivelyA. 2, 3B. – 2, 3C. 2, – 3D. 1, – 3Answer:ax2 – 5x + c = (2x + 1)(x – 3)= 2x2 – 6x + x – 3= 2x2 – 5x – 3On comparing the coefficients on both the sides, we get,a = 2 and c = – 3Question 7.If x + y = 10 and x – y = 2, then value of x isA. 4B. – 6C. – 4D. 6Answer:x + y = 10⇒ y = 10 – x substituting this value in x – y = 2We get,x – (10 – x) = 2⇒ 2x – 10 = 2⇒ 2x = 12⇒ x = 6Question 8.The solution of 2 – x <5 isA. x > – 3B. x < – 3C. x > 3D. x < 3Answer:2 – x <5Adding x on both the sides,2 <5 + x⇒ – 3 < x⇒ x > – 3

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