Section  1
Distribute the following expressions.
Ques 1: (x+ 2)(x 3)
Ans: (x + 2 )(x 3 )= x^{2} 3x + 2x  6 = x^{2} x  6
Ques 2: (2s+1)(s + 5)
Ans: (2s+ 1)(s+5) = 2s^{2} + 10s + s + 5 = 2s^{2}+ 11s + 5
Ques 3: (5 + a)(3 + a)
Ans: (5 + a)(3 + a) = 15 + 5a + 3a + a^{2} = a^{2} + 8 a + 15
Ques 4: (3  z)(z + 4)
Ans: (3  z)(z + 4) = 3z + 12  z^{2}  4z = z^{2}  z + 12
Section  2
Ques 5: x^{2}  2x = 0
Ans: x^{2} — 2x = 0
x(x — 2) = 0
x= 0
OR (x 2 ) = 0 > x = 2
Ques 6: z^{2} = 5 z
Ans:
Ques 7: y^{2} + 4y + 3 = 0
Ans:
Ques 8: y^{2} 11y + 30 = 0
Ans:
Ques 9: y^{2} + 3 y = 0
Ans:
Ques 10: y^{2} + 12y + 36 = 0
Ans:
Section  3
Simplify the following expressions.
Ques 11:
Ans: a + b_{1} The key to simplifying this expression is to recognize the special product:
After replacing the original numerator with (a + b)(a  b), we can cancel the (a  b) in the numerator with the (a  b) in the denominator:
Ques 12:
Ans:
Ques 13:
Ans:
Because we have a common factor in both terms of the numerator, we can divide that factor out in order to simplify further. This is often a useful move when we are asked to add or subtract exponents with the same base:
Ques 14:
Ans: It is tempting to expand the quadratic term in the numerator, but we should try to simplify first. Notice that none of the answer choices are fractions. Therefore, we need to look for a way to cancel (2t — 1) from the denominator. To make this task easier, enclose every (2t — 1) term in parentheses and then simplify. We can factor (2t  1) out of the numerator:
Alternatively, we can introduce a new variable x defined as x = 2t — 1. In general, it is not a good idea to introduce more variables than strictly necessary, but in this case the new variable can make it easier to see how the math works:
We can then simplify the expression in terms of x:
And we then finish by replacing x with 2t  1
x + 1 = (2t — 1) + 1 = 2t
Section  4
Simplify the following expressions.
Ques 15:
(A) 5  x (B) x  5 (C) x + 5
Ans: We can make this problem a lot simpler if we begin by factoring 3 out of both the nu
merator and the denominator:
The answers are not fractions, so well have to get rid of the denominator. Factor the numerator:
Ques 16:
Ans: This might seem nearly impossible to factor down until we take out the common term: ab. Then, we are left with one of our familiar special products in the denominator:
Ques 17:
Ans: There's no obvious way to proceed through this question. The best bet is to try to
simplify before multiplying. Notice that x^{3} can be factored out of the numerator and x^{2} can be factored out of the denominator:
Notice now that we can cancel x^{2}, from the fraction. Also, we have (x^{2}  1) in the numerator, which we can factor:
We don’t have a match with any of the answer choices yet. None of the numerators in the answer choices have parenthetical expressions. We should multiply the numerator:
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