Class 8 Factorization Exercise 7.9

\begin{array}{l} \text {Factorize each of the following quadratic polynomials by using the method of completing the square: } \\ \text {Q1. } \quad p^{2} + 6p + 8 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ p^{2} + 6p + 8 \\ \text {Here, Coefficient of } p^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } p \\\therefore \quad p^{2}+6p+8 = p^{2}+6p+3^{2}-3^{2}+8 \quad [\text {Adding and subtracting } (\frac{6}{2})^{2} = 3^{2}] \\=(p^{2}+ 2 \times 3 \times p + 3^{2} ) – 9 + 8 \\ =(p+3)^{2}-1^{2} \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =(p+3+1)(p+3-1) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(p+4)(p+2)\end{array}\begin{array}{l} \text {Q2. } \quad q^{2} -10q + 21 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ q^{2} -10q + 21 \\\text {Here, Coefficient of } q^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } q \\\therefore \quad q^{2} -10q + 21 = q^{2} -10q + 5^{2} – 5^{2} + 21 \quad [\text {Adding and subtracting } (\frac{10}{2})^{2} = 5^{2}] \\=(q^{2} – 2 \times 5 \times q + 5^{2} ) – 25 + 21 \\ =(q – 5)^{2}- 4 \quad [\text {Using : }(a-b)^{2} = a^{2} + b^{2} – 2ab ] \\ =(q – 5)^{2}- 2^{2} \\ =(q-5+2)(q-5-2) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(q-3)(q-7)\end{array}\begin{array}{l} \text {Q3. } \quad 4 y^{2} + 12y + 5 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ 4 y^{2} + 12y + 5 \\ \text {Coefficient of } y^{2} \text { is } 4. \\ \text {So, divide and multiply the expression by 4 so that coefficient of } y^{2} \text { will be 1.} \\4(y^{2} + \frac {12}{4}y + \frac {5}{4}) \\ =4(y^{2} +3y + \frac {5}{4}) \\\text {now add and subtract square of half of coefficient of } y \\\therefore \quad 4(y^{2} +3y + \frac {5}{4}) = 4[y^{2} +3y + \frac{3}{2})^{2} – \frac{3}{2})^{2} + \frac {5}{4}] \quad [\text {Adding and subtracting } (\frac{3}{2})^{2}] \\=4[y^{2} +2 \times \frac{3}{2} \times y + (\frac{3}{2})^{2}] – (\frac{3}{2})^{2} + \frac {5}{4} \\ =4[(y+\frac{3}{2})^{2} – \frac{9}{4} +\frac {5}{4}] \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =4[(y+\frac{3}{2})^{2} – 1] \\=4[(y+\frac{3}{2} + 1)(y+\frac{3}{2} – 1)] \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\=4[(y+\frac{5}{2})(y+\frac{1}{2})] \\ =4 \times \frac {(2y+5)}{2} \times \frac {(2y+1)}{2} \\ =(2y+5)(2y+1)\\ \end{array}\begin{array}{l} \text {Q4. } \quad p^{2} + 6p -16 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ p^{2} + 6p -16 \\\text {Here, Coefficient of } p^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } p \\\therefore \quad p^{2} + 6p -16 = p^{2}+6p+3^{2} -3^{2} – 16 \quad [\text {Adding and subtracting } (\frac{6}{2})^{2} = 3^{2}] \\=(p^{2}+ 2 \times 3 \times p + 3^{2} ) – 9 -16\\ =(p+3)^{2}-25 \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =(p+3)^{2}-5^{2} \\ =(p+3+5)(p+3-5) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(p+8)(p-2) \end{array}\begin{array}{l} \text {Q5. } \quad x^{2} + 12x + 20 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ x^{2} + 12x + 20 \\\text {Here, Coefficient of } x^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } x \\\therefore \quad x^{2} + 12x + 20 = x^{2}+ 12x + 6^{2} -6^{2} +20 \quad [\text {Adding and subtracting } (\frac{12}{2})^{2} = 6^{2}] \\=(x^{2}+ 2 \times 6 \times x + 6^{2} ) – 36 + 20 \\ =(x+6)^{2}-16 \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =(x+6)^{2}-4^{2} \\ =(x+6+4)(x+6-4) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(x+10)(x+2) \end{array}\begin{array}{l} \text {Q6. } \quad a^{2} – 14a – 51 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ a^{2} – 14a – 51 \\\text {Here, Coefficient of } a^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } a \\\therefore \quad a^{2} – 14a – 51 = a^{2} – 14a + 7^{2} -7^{2} -51 \quad [\text {Adding and subtracting } (\frac{14}{2})^{2} = 7^{2}] \\=(a^{2} – 2 \times 7 \times a + 7^{2} ) – 49 -51 \\ =(a-7)^{2}-100 \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =(a-7)^{2}-10^{2} \\ =(a-7+10)(a-7-10) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(a+3)(a -17) \end{array}\begin{array}{l} \text {Q7. } \quad a^{2} +2a – 3 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ a^{2} +2a – 3 \\\text {Here, Coefficient of } a^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } a \\\therefore \quad a^{2} +2a – 3 = a^{2} +2a + 1 -1 -3 \quad [\text {Adding and subtracting } (\frac{2}{2})^{2} = 1^{2} = 1] \\=(a^{2} + 2 \times 1 \times a + 1^{2} ) – 1 -3 \\ =(a+1)^{2}-4 \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =(a+1)^{2}-2^{2} \\ =(a+1+2)(a+1-2) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(a+3)(a-1) \end{array}\begin{array}{l} \text {Q8. } \quad 4 x^{2} – 12y + 5 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ 4 x^{2} – 12y + 5 \\ \text {Coefficient of } x^{2} \text { is } 4. \\ \text {So, divide and multiply the expression by 4 so that coefficient of } x^{2} \text { will be 1.} \\4(x^{2} – \frac {12}{4}x + \frac {5}{4}) \\ =4(x^{2} – 3x + \frac {5}{4}) \\\text {now add and subtract square of half of coefficient of } x \\\therefore \quad 4(x^{2} – 3x + \frac {5}{4}) = 4[x^{2} -3x + \frac{3}{2})^{2} – \frac{3}{2})^{2} + \frac {5}{4}] \quad [\text {Adding and subtracting } (\frac{3}{2})^{2}] \\=4[x^{2} – 2 \times \frac{3}{2} \times x + (\frac{3}{2})^{2}] – (\frac{3}{2})^{2} + \frac {5}{4} \\ =4[(x – \frac{3}{2})^{2} – \frac{9}{4} +\frac {5}{4}] \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =4[(x – \frac{3}{2})^{2} – 1] \\=4[(x-\frac{3}{2} + 1)(x-\frac{3}{2} – 1)] \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\=4[(x-\frac{1}{2})(x-\frac{5}{2})] \\ =4 \times \frac {(2x-1)}{2} \times \frac {(2x-5)}{2} \\ =(2x -5)(2x -1)\\ \end{array}\begin{array}{l} \text {Q9. } \quad y^{2} – 7y + 12 \\ \end{array}\begin{array}{l} \text {Sol. We have } \\ y^{2} – 7y + 12 \\\text {Here, Coefficient of } y^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } y \\\therefore \quad y^{2} – 7y + 12 \\ = y^{2} – 7y + (\frac{7}{2})^{2} -(\frac{7}{2})^{2} + 12 \quad [\text {Adding and subtracting } (\frac{7}{2})^{2} ] \\=(y^{2} – 2 \times \frac{7}{2} \times y + (\frac{7}{2})^{2} ) – \frac{49}{4} +12 \\ =(y-\frac{7}{2})^{2}- \frac{1}{4} \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =(y-\frac{7}{2})^{2}-(\frac{1}{2})^{2} \\ =(y-\frac{7}{2}+ \frac{1}{2})(y-\frac{7}{2} – \frac{1}{2}) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(y-\frac{6}{2})(y-\frac{8}{2}) \\ =(y-3)(y-4) \end{array}\begin{array}{l} \text {Q10. } \quad z^{2} -4z – 12\\ \end{array}\begin{array}{l} \text {Sol. We have } \\ z^{2} -4z – 12\\\text {Here, Coefficient of } z^{2} \text { is } 1. \\ \text {So, we add and subtract square of half of coefficient of } z \\\therefore \quad z^{2} -4z – 12 = z^{2} -4z + 2^{2} -2^{2} -12 \quad [\text {Adding and subtracting } (\frac{4}{2})^{2} = 2^{2} ] \\=(z^{2} – 2 \times 2 \times z + 2^{2} ) – 4 -12 \\ =(z-2)^{2} – 16 \quad [\text {Using : }(a+b)^{2} = a^{2} + b^{2} + 2ab ] \\ =(z-2)^{2}-4^{2} \\ =(z-2+4)(z-2-4) \quad [ \text {Using : } a^{2} – b^{2} = (a+b)(a-b) ] \\ =(z+2)(z-6) \end{array}