Factorization Of Algebraic Expressions
Today at school we have made rectangles and squares of different sizes using coloured cardboards.
Papia and Tathagaiha stuck those figures on a long thick chart paper and wrote the algebraic expressions for their area below the corresponding figures.
We all have planned to find out the length of each side of a rectangle or a square by fractions of each algebraic expression.
Question 1. On the school blackboard I resolve \(49 x^2+70 x y+25 y^2\)
Solution:
\(49 x^2+70 x y+25 y^2=7 x^2+2 x 7 x \times 5 y+5 y^2\)= \((7 x+5 y)^2\)
I got \(49 x^2+70 x y+25 y^2=(9 x+5 y) x(9 x+5 y)\)
The length of each side of the square = 7x + 5y unit.
Question 2. It is seen on the factorization of \(\left(81 a^2-72 a b+16 b^2\right)\)
The length of each side of the square = 9a – 4b
Read and Learn More WBBSE Solutions For Class 8 Maths
Question 3. Rehana factorized \(\left(64 m^2-121 n^2\right)\)
Solution:
She got, \(\left(64 m^2-121 n^2\right)\) = (8m + 11 n) x (8m – 11 n)
The length of the rectangle = 8m + 11n unit and the length of other side is units (8m – 11n).
Question 4. But Siraz factorized [/latex]\left(125 a^3+8 b^3\right)[/latex]
Solution:
Given
\(125 a^3+8 b^3=5 a^3+2 b^3\)= \((5 a+2 b)\left\{(5 a)^2-(5 a) \times(2 b)+4 b^2\right\}\)
[/latex]\left(125 a^3+8 b^3\right)[/latex] = (5a + 2b) (25a2 – 10ab + 4b2)[/latex][/latex]
Question 5. Utpal factorized \(\left(27 x^3-343 y^3\right)\)
Solution:
Given
\(27 x^3-343 y^3=3 x^3-7 y^3\)\(\left(27 x^3-343 y^3\right)\) = \((3 x-7 y)\left(9 x^2+21 x y+49 y^2\right)\)
I factorize \(\left(x^2+7 x+12\right)\) but how shall I write down \(\left(x^2+7 x+12\right)\) as the product of two algebraic expressions ?
Question 6. At first let us write ail known identities which help to factorize:
Solution:
\((a+b)^2=a^2+2 a b+b^2\) ………………………….. (1)
\((a-b)^2=a 2-2 a b+b^2\) ……………………………(2)
\(a^2-b^2=(a+b) \times(a-b)\) ………………………… (3)
\(a^3+b^3=(a+b) \times\left(a^2-a b+b^2\right)\) ………………….(4)
\(a^3-b^3=(a-b) \times\left(a^2+a b+b^2\right)\) …………………(5)
\(x^2+(a+b) x+a b=(x+a)(x+b)\) ………………….(6)
Question 7. I try to factorize the algebraic expression \(\left(x^2+7 x-18\right)\) with the help of identity no. 4
Solution:
\(x^2+7 x-18\)There a + b = 7 and a × b = -18
-18 = 1 x(-18)=(-1 )x(18)=(-2)x9=2x(-9)=(-3)x6=3x(-6) and 7=9+(-2)
∴ Here a = 9 and b = -2
∴ I get from the identity no. 4
\(x^2+7 x-18=(x+9)\{x+(-2)\}\)= (x+9)(x-2)
I get by factorizing
\(x^2+7 x-18=x^2+(9-2) x-18\)= \(x^2+9 x-2 x-18\)
= x (x+9)-2 (x+9) = (x+9)(x-2)
Question 8. Let’s try to factorize \(\left(a^2-11 a+30\right) \text { and }\left(m^2-4 m-12\right) \cdot a^2-11 a+30\)
Solution:
30= 1×30=2×15=3×10=5×6
30=5×6, 11=5+6
1. \(m^2-4 m-12\)
Solution:
-12=(-1)×12= 1×(-12) = (-2)×6
= 2×(-6)= 3×(-4)= -3×4
-12 =(-2)×6, 4 =6+(-2)
2. \(a^2-11 a+30\)
Solution:
= \(a^2-(5+6) a+30\)
= \(a^2-5 a-6 a+30\)
= a (a-5)-6(a-5)
= (a-5)(a-6)
\(a^2-11 a+30\) = (a-5)(a-6)
3. \(m^2-4 m-12\)
Solution:
= \(m^2-(6-2) m-12\)
= \(m^2-6 m+2 m-12\)
= m(m-6) +2(m-6)
= (m-6)(m+2)
∴ We will find out two numbers a and b to resolve the binomial expression \(x^2+p x+q\)(the highest power of the variable in the algebraic expression is 2) into factors so that
a+b = p and a x b =q
∴ In that case the algebraic expression will be
\(x^2+(a+b) x+a b\)=\(x^2+a x+b x+a b\)
= x (x+a) + b (x+a)
= (x+a) (x+b)
= (m-6)(m+2) = (x+a) (x+b)
Question 9. I try to resolve \(\left(x^2-x-20\right) \text { and }\left(b^2-10 b+16\right)\).
1. \(x^2-x-20\)
Solution:
-20 =5×(-4)
and 1= 5-4
2. \(b^2-10 b+16\)
Solution:
16= 8 × 2
and 10 = 8 + 2
3. \(x^2-x-20\)
Solution:
= \(x^2-(5-4) x-20\)
= \(x^2-5 x+4 x-20\)
= x(x-5)+4(x-5)
= (x-5)(x+4)
\(x^2-x-20\) = (x-5)(x+4)
4. \(b^2-10 b+16\)
Solution:
= \(b^2-(8+2) b+16\)
= \(b^2-8 b-2 b+16\)
= b(b-8)-2(b-8)
= (b-8)(b-2)
\(b^2-10 b+16\) = (b-8)(b-2)
Question 11. We factorize (x2+ 20xy – 96y2) and (1-5x-36×2). Each term of (x2+20xy-96y2) has variable. At first eliminate the variable from the last term.
Solution:
Given
\(x^2+20 x y-96 y^2\)= \(\left\{\frac{x^2}{y^2}+\frac{20 x y}{y^2}-\frac{96 y^2}{y^2}\right\} x y^2\)
= \(\left\{\left(\frac{x}{y}\right)^2+20\left(\frac{x}{y}\right)-96\right\} x y^2\)
= \(\left(a^2+20 a-96\right) y\) [Let \(\frac{x}{y}\) = a]
= \(\left(a^2+24 a-4 a-96\right) y^2\) [96=24×4 and 20=24-4]
= \(\{a(a+24)-4(a+24)\} y^2\)
= \((a+24)(a-4) y^2\)
= \(\left(\frac{x}{y}+24\right)\left(\frac{x}{y}-4\right) y^2\) [Puting a=\(\frac{x}{y}\)
= \(\left(\frac{x+24 y}{y}\right)\left(\frac{x-4 y}{y}\right) y^2\)
= \(\frac{(x+24 y)(x-4 y)}{y^2} y^2\)
= (x + 24y) (x-4y)
Alternative method
= \(x^2+20 x y-96 y^2\)
= \(x^2+(24 y-4 y) x-24 y \times 4 y\)
= \(x^2+24 x y-4 x y-24 y \times 4 y\)
= x (x + 24y) – 4y (x + 24y)
= x + 24y × x- 4y
Question 12. We try to factorize \(\left(1-5 x-36 x^2\right)\).
Solution:
Given
\(\left(1-5 x-36 x^2\right)\).
= \(1-5 x-36 x^2\)
= \(1-(9-4) \times(-9) \times 4 x^2\)
= \(1-9 x+4 x-9 \times 4 \times x^2\)
= 1 (1-9x) + 4x(1-9x)
= (1-9x)(1+4x)
\(\left(1-5 x-36 x^2\right)\) = (1-9x)(1+4x)
Question 13. I try to factorize (x-1) (x+3) (x-2)(x-6) +96.
Solution:
Given
(x-1) (x+3) (x-2)(x-6) +96.
But how will we factorize and with the help of which identity
x-1, x+3, x-2 and x-6? Let’s find the sum of the coefficients of two terms containing x only will be the product of two pair of expression among the four expressions
– 1-2 = – 3 and + 3 – 6= – 3
Then, the coefficient of x is the product (x-1)(x-2) is -3 and that of (x+3)(x-6) is (-3).
Then, (x-1)(x+3)(x-2)(x-6)+96
= (x-1 )(x+3)(x-2)(x-6) + 96
= \(\left(x^2-x-2 x+2\right)\left(x^2+3 x-6 x-18\right)+96\)
= \(\left(x^2-3 x+2\right)\left(x^2-3 x-18\right)+96\)
= (a+2) (a-18) + 96
= \(a^2-2 a-18 a-36+96\)
= \(a^2-16 a+60\)
= \(a^2-(10+6) a+60\)
= \(a^2-10 a-6 a+60\)
= a (a=10) -6 (a=10)
= (a-10) (a-6)
= \(\left(x^2-3 x-10\right)\left(x^2-3 x-6\right)\) [Let a=x2-3x]
= \(\left(x^2-5 x+2 x-10\right)\left(x^2-3 x-6\right)\)
= \(\{x(x-5)+2(x-5)\}\left(x^2-3 x-6\right)\)
= \((x-5)(x+2)\left(x^2-3 x-6\right)\)
(x-1) (x+3) (x-2)(x-6) +96 = \((x-5)(x+2)\left(x^2-3 x-6\right)\)
Question 14. Let’s factorize the algebraic expression \(x^2+3 x-(p+5)(p+2)\).
Solution:
Given
\(x^2+3 x-(p+5)(p+2)\) \(x^2+3 x-(p+5)(p+2)\)= \(x^2+\{(p+5)-(p+2)\} x-(p+5)(p+2)\)
= \(x^2+(p+5) x-(p+2) x-(p+5)(p+2)\)
= x (x+p+5) – (p+2) (x+p+5)
= (x+p+5) (x-p-2)
\(x^2+3 x-(p+5)(p+2)\) = (x+p+5) (x-p-2)
Algebraic Expressions Exercise
Question 1. Comparing the following algebraic expressions with the identity \(x^2+(p+q) x+p q=(x+p)(x+q)\), let’s write the values of p and q and factorize them.
Algebraic expression:
Given
\(x^2+(p+q) x+p q=(x+p)(x+q)\)= \(x^2-8 x+15\)
= \(x^2-40 x-129\)
= \(m^2+19 m+60\)
= \(x^2-x-6\)
= \((a+b)^2-4(a+b)-12\)
= \((x-y)^2-x+y-2\)
Values of p & q:
p= -5, q= -3
p=3, q=-43
p=15, q=4
p=-3, q=2
p=-6, q=2
p=-2, q=1
Resolve into factors:
= (x-5) (x-3)
= (x+3) (x-43)
= (m+15) (m+4)
= (x-3) (x+2)
= (a+b-6) (a+b+2)
= (x-y-2) (x-y+1)
Question 2. Let’s resolve into factors :
1. \((a+b)^2-5 a-5 b+6\)
= \((a+b)^2-5 a-5 b+6\)
= \((a+b)^2-5(a-b)+6\)
Let a + b = x
∴ \(x^2-5 x+6\)
= \(x^2-(3+2) x+6\)
= \(x^2-3 x-2 x+6\)
= x(x-3)-2 (x-3)
= (x-3) (x-2)
Putting the value of x (a+b-3) (a+b-2),
2. \(\left(x^2-2 x\right)^2+5\left(x^2-2 x\right)-36\)
Solution:
\(\left(x^2-2 x\right)^2+5\left(x^2-2 x\right)-36\)Let \(x^2-2 x=a\)
∴ \(a^2+5 a-36\)
= \(a^2+(9-4) a-36\)
= \(a^2+9 a-4 a-36\)
= a (a+9) – 4 (a+9)
= (a+9) (a-4)
Putting the value of a, (x2 – 2x +9) (x2 – 2x – 4)[/latex][/latex]
3. (p2-3q2)2-16(p2-3q2) + 63[/latex][/latex]
Solution:
(p2-3q2)2-16(p2-3q2) + 63[/latex][/latex]
Let x p2 – 3q2 = x[/latex][/latex]
∴ x2-16x + 63[/latex][/latex]
= x2 – (9+7) x + 63[/latex][/latex]
= x2 – 9x – 7x + 63[/latex][/latex]
= x (x-9) -7 (x-9)
= (x-9) (x-7)
Putting the value of x, (p2 – 3q2-9)(p2-3q2-7)[/latex][/latex]
4. a4+4a2-5[/latex][/latex]
Solution:
a4+4a4-5[/latex][/latex]
= a4+ (5-1) a2-5[/latex][/latex]
= a4 + 5a2-a2-5[/latex][/latex]
= a2 (a2+5) – 1 (a2+5)[/latex][/latex]
= (a2+5) (a2-1) = (a2+5) (a+1) (a-1)[/latex][/latex]
a4+4a2-5[/latex][/latex] = (a2+5) (a2-1) = (a2+5) (a+1) (a-1)[/latex][/latex]
5. \(x^2 y^2+23 x y-420\)
Solution:
\(x^2 y^2+23 x y-420\)= \(x^2 y^2+(35-12) x y-420\)
= \(x^2 y^2+35 x y-12 x y-420\)
= xy (xy + 35) -12 (xy + 35)
= (xy + 35) (xy -12)
\(x^2 y^2+23 x y-420\) = (xy + 35) (xy -12)
6. \(x^4-7 x^2+12\)
Solution:
= \(x^4-7 x^2+12\)
= \(x^4-(4+3) x^2+12\)
= \(x^4-4 x^2-3 x^2+12\)
= \(x^2\left(x^2-4\right)-3\left(x^2-4\right)\)
= \(\left(x^2-4\right)\left(x^2-3\right)\)
= \((x+2)(x-2)\left(x^2-3\right)\)
\(x^4-7 x^2+12\) = \((x+2)(x-2)\left(x^2-3\right)\)
7. \(a^2+a b-12 b^2\)
Solution:
\(a^2+a b-12 b^2\)= \(a^2+(4-3) a b-12 b^2\)
= \(a^2+4 a b-3 a b-12 b^2\)
= a(a+4b) -3b (a+4b)
= (a+4b) (a-3b)
\(a^2+a b-12 b^2\) = (a+4b) (a-3b)
8. \(p^2+31 p q+108 q^2\)
Solution:
\(p^2+31 p q+108 q^2\)= \(p^2+(27+4) p q+108 q^2\)
= \(p^2+27 p q+4 p q+108 q^2\)
= p (p+27q) + 4q (p+27q)
= (p+27q) (p+4q)
\(p^2+31 p q+108 q^2\) = (p+27q) (p+4q)
9. \(a^6+3 a^3 b^3-40 b^6\)
Solution:
\(a^6+3 a^3 b^3-40 b^6\)= \(a^6+(8-5) a^3 b^3-40 b^6\)
= \(a^6+8 a^3 b^3-5 a^3 b^3-40 b^6\)
= \(a^3\left(a^3+8 b^3\right)-5 b^3\left(a^3+8 b^3\right)\)
= \(\left(a^3+8 b^3\right)\left(a^3-5 b^3\right)\)
= \(\left\{(a)^3+(2 b)^3\right\}\left(a^3-5 b^3\right)\)
= \((a+2 b)\left(a^2-2 a b+4 b^2\right)\left(a^3-5 b^3\right)\)
\(a^6+3 a^3 b^3-40 b^6\) = \((a+2 b)\left(a^2-2 a b+4 b^2\right)\left(a^3-5 b^3\right)\)
10. (x+1) (x+3) (x-4) (x-6) + 24 .
Solution:
(x+1) (x+3) (x-4) (x-6) + 24
= (x+1) (x-4) (x+3) (x-6) + 24
= \(\left(x^2-3 x-4\right)\left(x^2-3 x-18\right)+24\)
Let \(x^2-3 x=a\)
∴ (a-4) (a-18) + 24
= \(a^2-22 a+72+24\)
= \(a^2-22 a+96\)
= \(a^2-(16+6) a+96\)
= \(a^2-16 a-6 a+96\)
= a (a-16) -6 (a-16)
= (a-16) (a-6)
(x+1) (x+3) (x-4) (x-6) + 24 = (a-16) (a-6)
Putting the value of a, \(\left(x^2-3 x-16\right)\left(x^2-3 x-6\right)\)
11. \((x+1)(x+9)(x+5)^2+63\)
Solution:
\((x+1)(x+9)(x+5)^2+63\)= \(\left(x^2+10 x+9\right)\left(x^2+10 x+25\right)+36\)
Let \(x^2+10 x=a\)
∴ (a+9) (a+25) + 63
= \(a^2+34 a+225+63\)
= \(a^2+34 a+288\)
= \(a^2+(18+16) a+288\)
= \(a^2+18 a+16 a+288\)
= a (a+18 + 16(a+18)
= (a+18) (a+16)
Putting the value of a, \(\left(x^2+10 x+18\right)\left(x^2+10 x+16\right)\)
= \(\left(x^2+10 x+18\right)\left\{\left(x^2+(8+2) x+16\right\}\right.\)
= \(\left(x^2+10 x+18\right)(x 2+8 x+2 x+16)\)
= \(\left(x^2+10 x+18\right)\{x(x+8)+2(x+8)\}\)
= \(\left(x^2+10 x+18\right)(x+8)(x+2)\)
= \((x+2)(x+8)\left(x^2+10 x+18\right)\)
12. x(x+3) (x+6) (x+9) + 56
Solution:
= x(x+3) (x+6) (x+9) + 56
= x(x+9) (x+3) (x+6) + 56
= \(\left(x^2+9 x\right)\left(x^2+9 x+18\right)+56\)
∴ Let \(x^2+9 x=a\)
= a (a+18) + 56
= \(a^2+18 a+56\)
= \(a^2+(14+4) a+56\)
= \(a^2+14 a+4 a+56\)
= a(a+14) +4 (a+14)
= (a+14) (a+4)
Putting the value of a,
\(\left(x^2+9 x+14\right)\left(x^2+9 x+4\right)\)= \(\left(x^2+7 x+2 x+14\right)\left(x^2+9 x+4\right)\)
= \(\{x(x+7)+2(x+7)\}\left(x^2+9 x+4\right)\)
= \((x+7)(x+2)\left(x^2+9 x+4\right)\)
13. \(x^2-2 a x+(a+b)(a-b)\)
Solution:
\(x^2-2 a x+(a+b)(a-b)\)= \(x^2-2 a x+a^2-b^2\)
= \((x-a)^2-(b)^2\)
= (x-a+b) (x-a-b)
\(x^2-2 a x+(a+b)(a-b)\) = (x-a+b) (x-a-b)
14. \(x^2-b x-(a+3 b)(a+2 b)\)
Solution:
\(x^2-\{(a+3 b)-(a+2 b)\} x-(a+3 b)(a+2 b)\)= \(x^2-(a+3 b) x+(a+2 b) x-(a+3 b)(a+2 b)\)
= x(x-a-3b) + (a + 2b) (x-a-3b)
= (x-a-3b) (x+a+2b)
\(x^2-b x-(a+3 b)(a+2 b)\) = (x-a-3b) (x+a+2b)
15. \((a+b)^2-5 a-5 b+6\)
Solution:
\((a+b)^2-5 a-5 b+6\)= \((a+b)^2-5(a+b)+6\)
Let a + b = x
∴ \(x^2-5 x+6\)
= \(x^2-(3+2) x+6\)
= \(x^2-3 x-2 x+6\)
= x (x-3) – 2 (x-3)
= (x-3) (x-2)
Putting the value of x, (a+b-3) (a+b-2)
\((a+b)^2-5 a-5 b+6\) = (a+b-3) (a+b-2)
16. \(x^2+4 a b x-\left(a^2-b^2\right)^2\)
Solution:
\(x^2+4 a b x-\left(a^2-b^2\right)^2\)= \(x^2+4 a b x-\{(a+b)(a-b)\}^2\)
= \(x^2+4 a b x-(a+b) 2(a-b)^2\)
= \(x^2+\left\{(a+b)^2-(a-b)^2\right\} x-(a+b)^2(a-b)^2\)
= \(x^2+(a+b)^2 x-(a-b) 2 x-(a+b)^2(a-b)^2\)
= \(x\left\{x+(a+b)^2\right\}-(a-b)^2\left\{x+(a+b)^2\right\}\)
= \(\left\{x+(a+b)^2\right\}\left\{x-(a-b)^2\right\}\)
= \(\left(x+a^2+2 a b+b^2\right)\left(x-a^2+2 a b-b^2\right)\)
\(x^2+4 a b x-\left(a^2-b^2\right)^2\) = \(\left(x+a^2+2 a b+b^2\right)\left(x-a^2+2 a b-b^2\right)\)
17. x2 – \(\left(a+\frac{1}{a}\right)\)x + 1
Solution:
= \(x^2-a x-\frac{x}{a}+\frac{a}{a}\)
= \(x(x-a)-\frac{1}{a}(x-a)=(x-a)\left(x-\frac{1}{a}\right)\)
x2 – \(\left(a+\frac{1}{a}\right)\)x + 1 = \(x(x-a)-\frac{1}{a}(x-a)=(x-a)\left(x-\frac{1}{a}\right)\)
18. \(x^6 y^6-9 x^3 y^3+8\)
Solution:
\(x^6 y^6-9 x^3 y^3+8\)= \(x^6 y^6-(8+1) x^3 y^3+8\)
= \(x^6 y^6-8 x^3 y^3-x^3 y^3+8\)
= \(x^3 y^3\left(x^3 y^3-8\right)-1\left(x^3 y^3-8\right)\)
= \(\left(x^3 y^3-8\right)\left(x^3 y^3-1\right)\)
= \(\left\{(x y)^3-(2)^3\left\{(x y)^3-(1)^3\right\}\right.\)
= \((x y-2)\left(x^2 y^2+2 x y+4\right)(x y-1)\left(x^2 y^2+x y+1\right)\)
\(x^6 y^6-9 x^3 y^3+8\) = \((x y-2)\left(x^2 y^2+2 x y+4\right)(x y-1)\left(x^2 y^2+x y+1\right)\)
Question 19. \(3 x^2+14 x+18\)
Solution:
= \(\frac{9 x^2+42 x+24}{3}\)
= \(\frac{(3 x)^2+14 \times 3 x+24}{3}\)
= \(\frac{y^2+14 y+24}{3}\) [Let 3x=y]
= \(\frac{y^2+12 y+2 y+24}{3}\)
= \( \frac{y(y+12)+2(y+12)}{3}\)
= \(\frac{(y+12)(y+2)}{3}\)
= \(\frac{(y+12)(y+2)}{3}\)
= \(\frac{(3 x+12)(3 x+2)}{3}\) [Putting y=3x,we get]
= \(\frac{3(x+4)(3 x+2)}{3}\)
= (x+4)(3x+2)
Alternatively we write,
\(3 x^2+14 x+8\)We find two numbers a and b such that a+b = 14 and a × b = 3 × 8 = 24
= 12×2 and 14= 12+2
= \(3 x^2+14 x+8\)
= \(3 x^2+(12+2) x+8\)
= \(3 x^2+12 x+2 x+8\)
= 3x (x+4) +2 (x+4)
= (x+4) (3x+2)
\(3 x^2+14 x+18\) = (x+4) (3x+2)
Question 20. Lets resolve \(6 x^2-x-15\) into factors and let’s write what are the possible lengths of the side (in units) of the yellow rectangular board.
Solution:
∴ -90 =10×(-9) and -1 = 10+(-9)
= \(6 x^2-x-15\)
= \(6 x^2-(10-9) x-15\)
= \(6 x^2-10 x+9 x-15\)
= 2x (3x-5)+3(3x-5)
= (3x-5) (2x + 3)
Question 21. Let’s express 2 algebraic expressions \(x^2+13 x-48 \text { and } 6 y^2-y-15\) as the difference between two squares. Now let’s try to resolve into factors with the help of the identity \(a^2-b^2=(a+b)(a+b)\).
Solution:
= \(x^2+13 x-48\)
= \( x^2+2 \cdot x \cdot \frac{13}{2}+\left(\frac{13}{2}\right)^2-\left(\frac{13}{2}\right)^2-48\)
= \(\left(x+\frac{13}{2}\right)^2-\frac{169}{4}-48\)
= \(\left(x+\frac{13}{2}\right)^2-\left(\frac{169}{4}+48\right)\)
= \( \left(x+\frac{13}{2}\right)^2-\frac{169+192}{4}\)
= \(\left(x+\frac{13}{2}\right)^2-\frac{361}{4}\)
= \(\left(x+\frac{13}{2}\right)^2-\left(\frac{19}{4}\right)^2\)
= \(\left(x+\frac{13}{2}+\frac{19}{2}\right)\left(x+\frac{13}{2}-\frac{19}{2}\right)\)
= \(\left(x+\frac{13+19}{2}\right)\left(x+\frac{13-19}{2}\right)\)
= (x+16) (x-3)
Factorize \(\left(x^2+13 x-48\right)\) by middle-term factor method.
\(x^2+13 x-48\)= \(x^2+16 x-3 x-48\)
= x (x+16) -3 (x+16)
= (x+16) (x-3)
Algebraic Expressions Exercise 13.2
Question 1. Let’s resolve into factors \(\left(a^2-a-72\right) \text { and }\left(2 x^2-x-1\right)\) by expressing the algebraic expressions as the difference of two squares.
Solution:
Given
\(\left(a^2-a-72\right) \text { and }\left(2 x^2-x-1\right)\)
\(a^2-a-72\)= \(a^2-(9-8) a-72\)
= \(a^2-9 a+8 a-72\)
= a (a-9) +8 (a-9)
= (a-9) (a+8)
\(2 x^2-x-1\)= \(2 x^2-(2-1) x-1\)
= \(2 x^2-2 x+x-1\)
= 2x (x-1) +1 (x-1)
= (x-1) (2x+1)
Question 2. Let’s resolve into factors —
1. \(2 a^2+5 a+2\)
Solution:
\(2 a^2+5 a+2\)= \(2 a^2+(4+1) a+2\)
= \(2 a^2+4 a+a+2\)
= 2a (a+2) +1 (a+2)
= (a+2) (2a+1).
\(2 a^2+5 a+2\) = (a+2) (2a+1).
2. \(3 x^2+14 x+8\)
Solution:
\(3 x^2+14 x+8\)= \(3 x^2+(12+2) x+8\)
= \(3 x^2+12 x+2 x+8\)
= 3x (x+4) +2 (x+4)
= (x+4) (3x+2)
\(3 x^2+14 x+8\) = (x+4) (3x+2)
3. \(2 m^2+7 m+6\)
Solution:
\(2 m^2+7 m+6\)= \(2 m^2+(4+3) m+6\)
= \(2 m^2+4 m+3 m+6\)
= 2m (m+2) +3 (m+2)
= (m+2) (2m+3)
\(2 m^2+7 m+6\) = (m+2) (2m+3)
4. \(6 x^2-x-15\)
Solution:
\(6 x^2-x-15\)= \(6 x^2-(10-9) x-15\)
= \(6 x^2-10 x+9 x-15\)
= 2x (3x – 5) +3 (3x – 5)
= (3x – 5) (2x + 3)
\(6 x^2-x-15\) = (3x – 5) (2x + 3)
5. \(9 r^2+r-8\)
Solution:
\(9 r^2+r-8\)= \(9 r^2+(9-8) r-8\)
= \(9 r^2+9 r-8 r-8\)
= 9r (r+1) -8 (r+1)
= (r+1) (9r-8)
\(9 r^2+r-8\) = (r+1) (9r-8)
6. \(6 m^2-11 m n-10 n^2\)
Solution:
\(6 m^2-11 m n-10 n^2\)= \(6 m^2-(15-4) m n-10 n^2\)
= \(6 m^2-15 m n+4 m n-10 n^2\)
= 3m (2m-5n) +2n (2m-5n)
= (2m-5n) (3m+2n)
\(6 m^2-11 m n-10 n^2\) = (2m-5n) (3m+2n)
7. \(7 x^2+48 x y-7 y^2\)
Solution:
\(7 x^2+48 x y-7 y^2\)= \(7 x^2+(49-1) x y-7 y^2\)
= \(7 x^2+49 x y-x y-7 y^2\)
= 7x (x+7y)-y(x+7y)
= (x+7-y) (7x-y)
\(7 x^2+48 x y-7 y^2\) = (x+7-y) (7x-y)
8. \(12+x-6 x^2\)
Solution:
\(12+x-6 x^2\)= \(12+(9-8) x-6 x^2\)
= \(12+9 x-8 x-6 x^2\)
= 3 (4+3x) – 2x (4+3x)
= (4+3x) (3-2x)
\(12+x-6 x^2\) = (4+3x) (3-2x)
9. \(6+5 a-6 a^2\)
Solution:
\(6+5 a-6 a^2\)= \(6+(9-4) a-6 a^2\)
= \(6+9 a-4 a-6 a^2\)
= 3 (2+3a)-2a (2+3a)
= (2+3a) (3-2a)
\(6+5 a-6 a^2\) = (2+3a) (3-2a)
10. \(6 x^2-13 x+6\)
Solution:
\(6 x^2-13 x+6\)= \(6 x^2-(9+4) x+6\)
= 3x (2x-3) -2 (2x-3)
= (2x-3) (3x-2)
\(6 x^2-13 x+6\) = (2x-3) (3x-2)
11. \(99 a^2-202 a b+99 b^2\)
Solution:
\(99 a^2-202 a b+99 b^2\)= \(99 a^2-(121+81) a b+99 b^2\)
= \(99 a^2-121 a b-81 a b+99 b^2\)
= 11 a (9a-11 b)-9b(9a-11 b)
= (9a-11b) (11a-9b)
\(99 a^2-202 a b+99 b^2\) = (9a-11b) (11a-9b)
12. \(2 a^6-13 a^3-24\)
Solution:
\(2 a^6-13 a^3-24\)= \(2 a^6-(16-3) a^3-24\)
= \(2 a^6-16 a 3+3 a^3-24\)
= \(2 a 3\left(a^3-8\right)+3\left(a^3-8\right)\)
= \(\left(a^3-8\right)\left(2 a^3+3\right)\)
= \(\left\{(a)^3-(2) 3\right\}\left(2 a^3+3\right)\)
= \((a-2)\left(a^2+2 a+4\right)\left(2 a^3+3\right)\)
\(2 a^6-13 a^3-24\)= \((a-2)\left(a^2+2 a+4\right)\left(2 a^3+3\right)\)
13. \(8 a^4+2 a^2-45\)
Solution:
\(8 a^4+2 a^2-45\)= \(8 a^4+(20-18) a^2-45\)
= \(8 a^4+20 a^2-18 a^2-45\)
= \(4 a^2\left(2 a^2+5\right)-9\left(2 a^2+5\right)\)
= \(\left(2 a^2+5\right)(4 a 2-9)\)
= \(\left(2 \mathrm{a}^2+5\right)\left\{(2 \mathrm{a})^2-(3)^2\right\}\)
= \(\left(2 a^2+5\right)(2 a+3)(2 a-3)\)
\(8 a^4+2 a^2-45\) = \(\left(2 a^2+5\right)(2 a+3)(2 a-3)\)
14. \(6(x-y)^2-x+y-15\)
Solution:
[/latex]6(x-y)^2-x+y-15[/latex]
= \(6(x-y)^2-(x-y)-15\)
Let x-y=a
∴ \(6 a^2-a-15\)
= \(6 a^2-(10-9) a-15\)
= \(6 a^2-10 a+9 a-15\)
= 2a (3a – 5) + 3 (3a – 5)
= (3a – 5) (2a + 3)
Putting the value of a, (3x-3y-5) (2x-2y+3)
\(6(x-y)^2-x+y-15\) = (3x-3y-5) (2x-2y+3)
15. \(3(a+b)^2-2 a-2 b-8\)
Solution.
\(3(a+b)^2-2 a-2 b-8\)= \(3(a+b)^2-2(a+b)-8\)
Let a + b = x
∴ \(3 x^2-2 x-8\)
= \(3 x^2-(6-4) x-8\)
= \(3 x^2-6 x+4 x-8\)
= 3x (x-2) +4 (x-2)
= (x-2) (3x+4)
Putting the value of x, (a+b-2) (3a+3b+4)
\(3(a+b)^2-2 a-2 b-8\) = (a+b-2) (3a+3b+4)
16. \(6(a+b)^2+5\left(a^2-b^2\right)-6(a-b)^2\)
Solution:
\(6(a+b)^2+5\left(a^2-b^2\right)-6(a-b)^2\)= \(6(a+b)^2+5(a+b)(a-b)-6(a-b)^2\)
Let a+b=x and a-b=y
∴ \(6 x^2+5 x y-6 y^2\)
= \(6 x^2+(9-4) x y-6 y^2\)
= \(6 x^2+9 x y-4 x y-6 y^2\)
= 3x (2x+3y) -2y (2x+3y)
= (2x+3y) (3x-2y)
Putting the values x and y, (2a+2b+3a-3b) (3a+3b-2a+2b) = (5a-b) (a+5b)
\(6(a+b)^2+5\left(a^2-b^2\right)-6(a-b)^2\)= (5a-b) (a+5b)
Question 3. Let’s resolve the following algebraic expressions into factors by expressing them as the difference of two squares.
1. \(x^2-2 x-3\)
Solution:
\(x^2-2 x-3\)= \(x^2-2 x+1-4\)
= \((x)^2-2 \times 1+(1) 2-(2)^2\)
= \((x-1)^2-(2)^2\)
= (x-1+2) (x-1-2)
= (x+1) (X-3)
\(x^2-2 x-3\) = (x+1) (X-3)
2. \(x^2+5 x+6\)
Solution:
\(x^2+5 x+6\)= \((x)^2+2 \cdot x \cdot \frac{5}{2}+\left(\frac{5}{2}\right)^2+6-\left(\frac{5}{2}\right)^2 \)
= \(\left(x+\frac{5}{2}\right)^2+6-\frac{25}{4}\)
= \(\left(x+\frac{5}{2}\right)^2+\frac{24-25}{4}\)
= \(\left(x+\frac{5}{2}\right)^2-\frac{1}{4}\)
= \(\left(x+\frac{5}{2}\right)^2-\left(\frac{1}{2}\right)^2\)
= \( \left(x+\frac{5}{2}+\frac{1}{2}\right)\left(x+\frac{5}{2}-\frac{1}{2}\right)\)
= (x+3) (x+2)
\(x^2+5 x+6\) = (x+3) (x+2)
3. \(3 x^2-7 x-6\)
Solution:
\(3 x^2-7 x-6\)= \(3\left(x^2-\frac{7}{3} x-2\right)\)
= \( \left\{(x)^2-2 \cdot x \cdot \frac{7}{6}+\left(\frac{7}{6}\right)^2-\left(\frac{7}{6}\right)^2-2\right\}\)
= \(3\left\{\left(x-\frac{7}{6}\right)^2-\left(\frac{49}{36}+\frac{2}{1}\right)\right\}\)
\(=3\left\{\left(x-\frac{7}{6}\right)^2-\frac{121}{36}\right\}\) \(=3\left\{\left(x-\frac{7}{6}\right)^2-\left(\frac{11}{6}\right)^2\right\}\)
= \(3\left(x-\frac{7}{6}+\frac{11}{6}\right)\left(x-\frac{7}{6}-\frac{11}{6}\right) \)
= \(3\left(x+\frac{4}{6}\right)(x-3)\)
= (3x+2) (x-3)
\(3 x^2-7 x-6\) = (3x+2) (x-3)
4. \(3 a^2-2 a-5\)
Solution:
= \(3\left(a^2-\frac{2}{3} a-\frac{5}{3}\right)\)
= \(3\left\{(a)^2-2 \cdot a \cdot \frac{1}{3}+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2-\frac{5}{3}\right\}\)
= \(3\left\{\left(a-\frac{1}{3}\right)^2-\frac{1}{9}-\frac{5}{3}\right\}\)
= \(3\left\{\left(a-\frac{1}{3}\right)^2-\left(\frac{1+15}{9}\right)\right\}\)
= \(3\left\{\left(a-\frac{1}{3}\right)^2-\frac{16}{9}\right\}\)
= \(3\left\{\left(a-\frac{1}{3}\right)^2-\left(\frac{4}{3}\right)^2\right\}\)
= \(3\left(a-\frac{1}{3}+\frac{4}{3}\right)\left(a-\frac{1}{3}-\frac{4}{3}\right)\)
= \(3(a+1)\left(a-\frac{5}{3}\right)\)
= (a+1) (3a-5)
\(3 a^2-2 a-5\) = (a+1) (3a-5)
Question 4. Let’s resolve into factors
1. \(a x^2+\left(a^2+1\right) x+a\)
Solution:
\(a x^2+\left(a^2+1\right) x+a\)= \(a x^2+a^2 x+x+a\)
= ax (x + a) + 1 (x + a)
= (x + a) (ax + 1)
\(a x^2+\left(a^2+1\right) x+a\) = (x + a) (ax + 1)
2. \(x^2+2 a x+(a+b)(a-b)\)
Solution:
\(x^2+2 a x+(a+b)(a-b)\)= \(x^2+2 a x+a^2-b^2\)
= \((x+a)^2-(b)^2\)
= (x + a + b) (x + a – b)
\(x^2+2 a x+(a+b)(a-b)\) = (x + a + b) (x + a – b)
3. \(a x^2-\left(a^2+1\right) x+a\)
\(a x^2-\left(a^2+1\right) x+a\)= \(a x^2-a^2 x-x+a\)
= ax (x – a) -1 (x – a)
= (x – a) (ax – 1)
\(a x^2-\left(a^2+1\right) x+a\) = (x – a) (ax – 1)
4. \(a x^2+\left(a^2-1\right) x-a\)
Solution:
\(a x^2+\left(a^2-1\right) x-a\)= \(a x^2+a^2 x-x-a\)
= ax (x + a) – 1 (x + a)
= (x + a) (ax – 1)
\(a x^2+\left(a^2-1\right) x-a\) = (x + a) (ax – 1)
5. \(a x^2-\left(a^2-2\right) x-2 a\)
Solution:
\(a x^2-\left(a^2-2\right) x-2 a\)= \(a x^2-a^2 x+2 x-2 a\)
= ax (x – a) + 2 (x – a)
= (x – a) (ax + 2)
\(a x^2-\left(a^2-2\right) x-2 a\) = (x – a) (ax + 2)
6. \(a^2+1-\frac{6}{a^2}\)
Solution:
= \(a^2+1-\frac{6}{a^2}\)
= \(a^2+3-2-\frac{6}{a^2}\)
= \(a\left(a+\frac{3}{a}\right)-\frac{2}{a}\left(a+\frac{3}{a}\right)\)
= \(\left(a+\frac{3}{a}\right)\left(a-\frac{2}{a}\right)\)
\(a^2+1-\frac{6}{a^2}\) = \(\left(a+\frac{3}{a}\right)\left(a-\frac{2}{a}\right)\)