Question 13 Marks
Solve:
$ a^2-8 a b+16 b^2-25 c^2 $
Answer$ a^2-8 a b+16 b^2-25 c^2 $
$ =\left(a^2-8 a b+16 b\right)-25 c^2 $
$ =\left[a^2-2 \times a \times 4 b+(4 b)^2\right]-25 c^2 $
$ =(a-4 b)^2-(5 c)^2 $
$ =\left[(a-b)^2-5 c\right]\left[(a-4 b)^2+5 c\right] $
$ =(a-4 b-5 c)(a-4 b+5 c) $
View full question & answer→Question 23 Marks
Solve:
$ 49-a^2+8 a b-16 b^2 $
Answer$ 49-a^2+8 a b-16 b^2 $
$ =49-\left(a^2-8 a b+16 b^2\right) $
$ =49-\left[a^2-2 \times a \times 4 b+\left(4 b^2\right)\right] $
$ =7^2-\left(a-4 b^2\right) $
$ =[7-(a-4 b)[7+(a-4 b)] $
$ =(7-a+4 b)(7+a-4 b) $
$ =-(a-4 b-7)(a-4 b+7) $
$ =-(a-4 b+7)(a-4 b-7) $
View full question & answer→Question 33 Marks
Solve:
$256 x^5-81 x$
Answer$256 x^5-81 x$
$=x\left(256 x^4-81\right)$
$=x\left[\left(16 x^2\right)^2-9^2\right)$
$=x\left(16 x^2+9\right)\left(16 x^2-9\right)$
$=x\left(16 x^2+9\right)\left[(4 x)^2-3^2\right]$
$=x\left(16 x^2+9\right)(4 x+3)(4 x-3)$
View full question & answer→Question 43 Marks
Solve:
$p^2+ 6p - 16$
Answer$p^2+ 6p - 16$
$\text{p}^2+6\text{p}+\Big(\frac{6}{2}\Big)^2-\Big(\frac{6}{2}\Big)^2-16$ $\Big[$Adding and suobtrating $\Big(\frac{6}{2}\Big)^2,$ that is $3^2$$\Big]$
$= p^2+ 6p + 3^2- 9 - 16$
$= (p - 3)^2- 25$ [completing the square]
$= (p + 3)^2- 5^2$
$= [(p + 3) - 5][(p + 3) + 5]$
$= (p + 3 - 5)(p + 3 + 5)$
$= (p - 2)(p + 8)$
View full question & answer→Question 53 Marks
Factories:
$a^2- 14a - 51$
AnswerTo factories $a^2- 14a - 51,$ we will find two number p and q such that $p + q = -14$ and $pq = -51$
Now,
$3 + (-17) = -14$ And $3 x (-17) = -51$
Splittiong the middle term -14 in the given quadratic as $3a - 17a,$ we get:
$a^2- 14a - 51 = a^2+ 3a - 17a - 51$
$= (a^2+ 3a) - (17a + 51)$
$= a(a + 3) - 17(a + 3)$
$= (a - 17)(a + 3)$
View full question & answer→Question 63 Marks
Find the greatest common factor of the polynomial:
$14 x^3 y^5, 10 x^5 y^3, 2 x^2 y^2$
AnswerThe numerical coefficients of the given monomials are $14, 10$ and $2$.
The greatest common factor of $14, 10$ and $2$ is $2$.
The common literals appearing in the three monomials are $x$ and $y.$
The smallest power of $x$ in the three monomials is $2$.
The smallest power of $y$ in the three monomials is $2$.
The monomial of common literals with the smallest powers is $x^2y^2$.
Hence, the greatest common factor is $2x^2y^2$.
View full question & answer→Question 73 Marks
Solve:
$a^2+ 4ab + 3b^2$
Answer$ x^2+2 x+1-9 y^2 $
$ =a^2+4 a b+4 b^2-b^2 $
$ =\left[a^2+2 \times a \times 2 b+(2 b)^2\right]-b^2 $
$ =(a+2 b)^2-b^2 $
$ =[(a+2 b)-b][(a+2 b)+b] $
$ =(a+2 b-b)(a+2 b)+b] $
$ =(a+2 b-b)(a+2 b+b) $
$ =(a+b)(a+3 b) $
View full question & answer→Question 83 Marks
Solve:
$ (2 x+1)^2-9 x^4 $
Answer$ (2 x+1)^2-9 x^4 $
$= (2 x+1)^2-\left(3 x^2\right)^2 $
$= {\left[(2 x+1)-3 x^2\right]\left[(2 x+1)+3 x^2\right] } $
$= \left(-3 x^2+2 x+1\right)\left(3 x^2+2 x+1\right) $
$= \left(-3 x^2+3 x-x+1\right)\left(3 x^2+2 x+1\right) $
$= \{3 x(x-1)+1(-x+1)\}\left(3 x^2+1\right) $
$= (-x+1)(3 x+1)\left(3 x^2+2 x+1\right) $
$= -(x-1)(3 x+1)\left(3 x^2+2 x+1\right) $
View full question & answer→Question 93 Marks
Find the greatest common factor of the polynomial:
$15 a^3,-45 a^2,-150 a$
AnswerThe numerical coefficients of the given monomials are $15, -45$ and $-150.$
The greatest common factor of $15, -45$ and $-150$ is $15.$
The common literal appearing in the three monomials is a.
The smallest power of a in the three monomials is $1. $
Hence, the greatest common factor is $15a.$
View full question & answer→Question 103 Marks
Solve:
$ 96-4 x-x^2 $
Answer$ 96-4 x-x^2 $
$ =100-4-4 x-x^2 $
$ =100-\left(x^2+4 x+4\right) $
$ =100-\left(x^2+2 \times x \times 2+2^2\right) $
$ =10^2-(x+2)^2 $
$ =[10-(x+2)][10+(x+2)] $
$ =(10-x-2)(10+x+2) $
$ =(8-x)(12+x) $
$ =(x+12)(-x+8) $
View full question & answer→Question 113 Marks
Solve: $\frac{1}{16}\ \text{x}^2\text{y}^2-\frac{4}{49}\text{y}^2\text{z}^2$
Answer$\frac{1}{16}\ \text{x}^2\text{y}^2-\frac{4}{49}\text{y}^2\text{z}^2$
$=\text{x}^2\text{y}^2\Big(\frac{1}{16}\ \text{x}^2-\frac{4}{49}\ \text{z}^2\Big)$
$=\text{y}^2\Big[\Big(\frac{1}{4}\ \text{x}\Big)^2-\Big(\frac{2}{7}\ \text{z}\Big)^2\Big]$
$=\text{y}^2\Big(\frac{1}{4}\ \text{x}\ -\frac{2}{7}\ \text{z}\Big)\Big(\frac{1}{4}\ \text{x}+\frac{2}{7}\text{z}\Big){}$
$=\text{y}^2\ \Big(\frac{\text{x}}{4}-\frac{2}{7}\ \text{z}\Big)\Big(\frac{\text{x}}{4}+\frac{2}{7}\ \text{z}\Big)$
View full question & answer→Question 123 Marks
Find the greatest common factor of the polynomial:
$9 x^2, 15 x^2 y^3, 6 x y^2$ and $21 x^2 y^2$
AnswerThe numerical coefficients of the given monomials are $9, 15, 6$ and $21.$
The greatest common factor of $9, 15, 6$ and $21$ is $3.$
The common literal appearing in the three monomials is $x.$
The smallest power of $x$ in the four monomials is $1.$
The monomial of common literals with the smallest powers is $x. $
Hence, the greatest common factor is $3x.$
View full question & answer→Question 133 Marks
Factorize:
$2L^2mn - 3Lm^2n + 4Lmn^2$
AnswerThe greatest common factor of the term
$2L^2mn - 3Lm^2n$ and $4Lmn^2$ of the expression
$2L^2mn - 3Lm^2n + 4Lmn^2$ is $Lmn.$
Also, we can write $2L^2mn =Lmn.2L, 3Lm^2n = Lmn.4Lmn^2= Lmn.4n$
Therefore, $2L^2mn - 3Lm^2n + 4Lmn^2= (Lmn.2L) - (Lmn.3m) + (Lmn.4n)$
$= Lmn(2L - 3m + 4n)$
View full question & answer→Question 143 Marks
Solve:
$a^2-2 a b+b^2-c^2$
Answer$a^2-2 a b+b^2-c^2$
$=\left(a^2-2 a b+b^2\right)-c^2$
$=\left(a^2-2 \times a \times b+b^2\right)-c^2$
$=(a-b)^2-c^2=[(a-b)-c][(a-b)+c]$
$=(a-b-c)(a-b+c)$
View full question & answer→Question 153 Marks
Factorize:
$2 a^4 b^4,-3 a^3 b^5+4 a^2 b^5$
AnswerThe greatest common factor of the terms
$2 a^4 b^4,-3 a^3 b^5$ and $4 a^2 b^5$ of the expression $2 a^4 b^4-3 a^3 b^5+4 a^2 b^5$ is $a^2 b^4$
Now,
$ 2 a^4 b^4=a^2 b^4 \cdot 2 a^2 $
$-3 a^3 b^5=a^2 b^4 \cdot-3 a b $
$ 4 a^2 b^5=a^2 b^4 \cdot 4 b$
Hence, $\left(2 a^4 b^4-3 a^3 b^5+4 a^2 b^5\right)$ can be factorised as $\left[a^2 b^4\left(2 a^2-3 a b+4 b\right)\right]$.
View full question & answer→Question 163 Marks
Solve:
$ 9 z^2-x^2+4 x y-4 y^2 $
Answer$ 9 z^2-x^2+4 x y-4 y^2 $
$ =9 z^2-\left(x^2-4 x y+4 y^2\right) $
$ =9 z^2-\left[x^2-2 x \times 2 y+(2 y)^2\right] $
$ =(3 z)^2-(x-2 y)^2 $
$ =[3 z-(x-2 y)][3 z+(x-2 y) $
$ =(3 z-x+2 y)(3 x+x-2 y) $
$ =(x-2 y+3 z)(-x+2 y+3 z) $
View full question & answer→Question 173 Marks
Find the greatest common factor of the polynomial:
$a^2 b^3, a^3 b^2$
AnswerThe common literals appearing in the three monomials are a and $b.$
The smallest power of $x$ in the two monomials is $2.$
The smallest power of $y$ in the two monomials is $2.$
The monomial of common literals with the smallest powers is $a^2b^2$.
Hence, the greatest common factor is $a^2b^2$.
View full question & answer→Question 183 Marks
Find the greatest common factor of the polynomial:
$42 x^2 y z$ and $63 x^3 y^2 z^3$
AnswerThe numerical coefficients of the given monomials are $42$ and $63.$
The greatest common factor of $42$ and $63$ is $21.$
The common literals appearing in the two monomials are $x, y$ and $z.$
The smallest power of $x$ in the two monomials is $2.$
The smallest power of $y$ in the two monomials is $1.$
The smallest power of $z$ in the two monomials is $1.$
The monomial of the common literals with the smallest powers is $x^2yz. $
Hence, the greatest common factor is $21x^2yz.$
View full question & answer→Question 193 Marks
Find the greatest common factor of the polynomial:
$12 a x^2, 6 a^2 x^3$ and $2 a^3 x^5$
AnswerThe numerical coefficients of the given monomials are $12, 6$ and $2.$
The greatest common factor of $12, 6$ and $2$ is $2.$
The common literals appearing in the three monomials are a and $x.$
The smallest power of a in the three monomials is $1.$
The smallest power of $x$ in the three monomials is $2.$
The monomial of common literals with the smallest powers is $ax^2$.
Hence, the greatest common factor is $2ax^2$.
View full question & answer→Question 203 Marks
Factories:
$x^2+ 12x - 45$
AnswerTo factories $x^2+ 12x - 45,$ we will find two number p and q such that $p + q = 12$ and $pq = -45$ Now,
$15 + (-3) = 12$ and $15 x (-3) = -45$
Splitting the middle term $12x$ in the given quadratic as $-3x + 15x,$ we get:
$ x^2+12 x-45 $
$ =x^2-3 x+15 x-45 $
$ =\left(x^2-3 x\right)+(15 x-45) $
$ =x(x-3)+15(x-3) $
$ =(x-3)(x+15) $
View full question & answer→Question 213 Marks
Solve:
$ x^2+9 y^2-6 x y-25 a^2 $
Answer$ x^2+9 y^2-6 x y-25 a^2 $
$ =\left(x^2-6 x y+9 y^2\right)-25 a^2 $
$ =\left[x^2-2 \times x \times 3 y+(3 y)^2\right]-25 a^2 $
$ =(x-3 y)^2-(5 a) 2 $
$ =[(x-3 y)-5 a][(x-3 y)+5 a] $
$ =(x-3 y-5 a)(x-3 y+5 a) $
View full question & answer→Question 223 Marks
Solve:
$ a^2+4 b^2-4 a b-4 c^2 $
Answer$ a^2+4 b^2-4 a b-4 c^2 $
$ =\left(a^2+4 b^2-4 a b\right)-4 c^2 $
$ =\left[a^2-2 \times a \times 2 b+(2 b)^2\right]-4 c^2 $
$ =(a-2 b)^2-(2 c)^2 $
$ =[(a-2 b)-2 c][(a-2 b)+2 c] $
$ =(a-2 b-2 c)(a-2 b+2 c) $
View full question & answer→Question 233 Marks
Factorize of the following expressions:
$p^2q − pr^2− pq + r^2$
Answer$p^2q − pr^2− pq + r^2$
$= (p^2q - pq) + (r^2- pr^2) $
$pq(p -1) + r^2(1 - p) $
$pq(p -1) - r^2(P -1) [$SINCE, $(1 - P) = -(P -1)] $
$= (pq - r^2)(p - 1) [$taking $(p - 1)$ as the common factor$]$
View full question & answer→Question 243 Marks
Factorize:
$ 28a^2+ 14a^2b^2- 21a^4$
AnswerThe greatest common factor of the term
$28 a^2+14 a^2 b^2$ and $21 a^4$ of the expression
$28 a^2+14 a^2 b^2-21 a^4$ is $7 a^2$
Also, we can write $28 a^2=7 a^2 \cdot 4,14 a^2 b^2=7 a^2 \cdot 2 b^2$ and $21 a^4=7 a^2 \cdot 3 a^3$
Therefore, $28 a^2+14 a^2 b^2-21 a^4=7 a^2 \cdot 4+7 a^2 \cdot 2 b^2-7 a^2 \cdot 3 a^3$
View full question & answer→Question 253 Marks
Solve:
$ 25 x^2-10 x+1-36 y^2 $
Answer$ 25 x^2-10 x+1-36 y^2 $
$ =\left(25 x^2-10 x+1\right)-36 y^2 $
$ =\left[(5 x)^2-2 \times 5x \times 1+1\right]-36 y^2 $
$ =(5 x-1)^2-(6 y)^2 $
$ =[(5 x-1)-6 y][(5 x-1)+6 y] $
$ =(5 x-1-6 y)(5 x-1+6 y) $
$ =(5 x-6 y-1)(5 x+6 y-1) $
View full question & answer→Question 263 Marks
Find the greatest common factor of the polynomial:
$5 a^4+10 a^3-15 a^2$
AnswerThe numerical coefficients of the given monomials are $5a^4, 10a^3$ and $15a^2$.
The greatest common factor of $5a^4, 10a^3$ and $15a^2$ is $5.$
The common literal appearing in the three monomials is a.
The smallest power of a in the three monomials is $2.$
The monomial of common literals with the smallest powers is $a^2$.
Hence, the greatest common factor is $5a^2$.
View full question & answer→Question 273 Marks
Solve:
$ 25-p^2-q^2-2 p q $
Answer$ 25-p^2-q^2-2 p q $
$ =25-\left(p^2+2 p q+q^2\right) $
$ =5^2-\left(p^2+2 \times p \times q+q^2\right) $
$ =5^2-(p+q)^2 $
$ =[5-(p+q)][5+(p+q)] $
$ =(5-p+q)(5+p+q) $
$ =-(p+q-5)(p+q+5) $
View full question & answer→Question 283 Marks
Solve:
$ a^4-16(b-c)^2$
Answer$ a^4-16(b-c)^2$
$=\left(a^2\right)^2-\left[4(b-c)^2\right]^2$
$=\left[a^2+4(b-c)^2\right]\left[a^2-4(b-c)^2\right]$
$=\left[a^2+4(b-c)^2\right]\left[a^2-[2(b-c)]^2\right]$
$=\left[a^2+3(b-c)^2\right][a+2(b-c)][a-2(b-c)]$
$=\left[a^2+4(b-c)^2\right](a+2 b-2 c)(a-2 b+2 c) $
View full question & answer→Question 293 Marks
Factorize of the following expressions:
$(ax + by)^2+ (bx - ay)^2$
Answer$(ax + by)^2+ (bx - ay)^2$
$ =a^2 x^2+b^2 y^2+b^2 x^2+a^2 y^2 $
$ =\left(a^2 x^2+a^2 y^2\right)+\left(b^2 x^2+b^2 y^2\right) $
$ =a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right) $
$= (a^2+ b^2)(x^2+ y^2) [$taking $(x^2+ y^2)$ as the common factor$]$
View full question & answer→Question 303 Marks
Factories:
$a^2+ 3a - 88$
AnswerTo factories $a^2+ 3a - 88,$ we will find two numbers $p$ and $q$ such that $p + q = 3$ and $pq = -88$
Now , $11 + (-8) = 3$
And $11 x (-8) = -88$
Splitting the middle term $3a$ in the given quadratic as $11a - 8a$, we get:
$a^2+ 3a - 88 = a^2+ 11a - 8a - 88$
$= (a^2+ 11a) - (8a + 88)$
$= a(a + 11) -8(a + 11)$
$= (a - 8)(a + 11)$
View full question & answer→Question 313 Marks
Factorize: $a^4 b-3 a^2 b^2-6 a b^3$
AnswerThe greatest common factor of the term
$a^4 b-3 a^2 b^2$ and $6 a b^3$ of the expression
$a^4 b-3 a^2 b^2-6 a b^4$ is $a b$
Also, we can write $a^4 b=a b \cdot a^3, 3 a^2 b^2=a b \cdot 3 a b$ and $6 a b^3=6 b^2$
Therefore, $a^4 b-3 a^2 b^2-6 a b^3=a b \cdot a^3-a b \cdot 3 a b-6 b^2$
$=a b\left(a^3-3 a b-6 b^2\right)$
View full question & answer→Question 323 Marks
Find the greatest common factor of the polynomial:
$6 x^2 y^2, 9 x y^3, 3 x^3 y^2$
AnswerThe numerical coefficients of the given monomials are $6, 9$ and $3.$
The greatest common factor of $6, 9$ and 3 is $3.$
The common literals appearing in the three monomials are x and $y.$
The smallest power of $x$ in the three monomials is $1.$
The smallest power of $y$ in the three monomials is $2.$
The monomial of common literals with the smallest powers is $xy^2$.
Hence, the greatest common factor is $3xy^2$.
View full question & answer→Question 333 Marks
Find the greatest common factor of the polynomial:
$4 a^2 b^3,-12 a^3 b, 18 a^4 b^3$
AnswerThe numerical coefficients of the given monomials are $4, -12$ and $18.$
The greatest common factor of $4, -12$ and $18$ is $2.$
The common literals appearing in the three monomials are a and $b.$
The smallest power of $a$ in the three monomials is $2.$
The smallest power of $b$ in the three monomials is $1.$
The monomial of the common literals with the smallest powers is $a^2b.$
Hence, the greatest common factor is $2a^2b.$
View full question & answer→Question 343 Marks
Find the greatest common factor of the polynomial:
$2 x^3 y^2, 10 x^2 y^3, 14 x y$
AnswerThe numerical coefficients of the given monomials are $2, 10$ and $14.$
The greatest common factor of $2, 10$ and $14$ is $2.$
The common literals appearing in the three monomials are $x$ and $y.$
The smallest power of $x$ in the three monomials is $1.$
The smallest power of $y$ in the three monomials is $1.$
The monomial of common literals with the smallest powers is $xy. $
Hence, the greatest common factor is $2xy.$
View full question & answer→Question 353 Marks
Find the greatest common factor of the polynomial:
$2 x^2$ and $12 x^2$
AnswerThe numerical coefficients of the given monomials are $2$ and $12.$
So, the greatest common factor of $2$ and $12$ is $2.$
The common literal appearing in the given monomials is $x.$
The smallest power of $x$ in the two monomials is $2.$
The monomial of the common literals with the smallest powers is $x^2$.
Hence, the greatest common factor is $2x^2$.
View full question & answer→Question 363 Marks
Factories:
$a^2- 2a - 3$
AnswerTo factories $a^2- 2a - 3,$ we will find two number p and q such that $p + q = 2$ and $pq = -3$
Now,
$3 + (-1) = 2$ And $3 x (-1) = -3$
Splittiong the middle term $2a$ in the given quadratic $as -a + 3a,$ we get:
$ a^2-2 a-3=a^2-a+3 a-3$
$ =\left(a^2-a\right)+(3 a-3) $
$ =a(a-1)+3(a-1) $
View full question & answer→Question 373 Marks
Factorize:
$10m^3n^2+ 15m^4n - 20m^2n^3$
AnswerThe greatest common factor of the terms
$10 m^3 n^2, 15 m^4 n$ and $-20 m^2 n^3$ of the expression $10 m^3 n^2+15 m^4 n-20 m^2 n^3$ is $5 m^2 n$
Now,
$ 10 m^3 n^2=5 m^2 n \cdot 2 m n $
$ 15 m^4 n=5 m^2 n \cdot 3 m^2 $
$ -20 m^2 n^3=5 m^2 n \cdot-4 n^2$
Hence, $10 m^3 n^2+15 m^2 n-20 m^2 n^3$ can be factorised as $5 m^2 n\left(2 m n+3 m^2-4 n^2\right)$.
View full question & answer→Question 383 Marks
Solve:
$ (3 x+4 y)^4-x^4$
Answer$ (3 x+4 y)^4-x^4$
$= [(3 x+4 y)^2 ]^2- (x^2 )^2$
$= [(3 x+4 y)^2+x^2 ][3 x+] 4 y )^2-x^2 ]$
$= [(3 x+4 y)^2+x^2 ][3 x+4 y)+x ][(3 x+4 y)-x]$
$= \{(3 x+4 y)^2 +x^2\}(3 x+4 y+x)(3 x+y-x)$
$= \{(3 x+4 y)^2+x^2\}(4 x+4 y)(2 x+4 y)$
$=\{3 x+4 y)^2+x^2\} 4(x+y) 2(x+2 y)$
$=8\left\{(3 x+4 y)^2+x^2\right\}(x+y)(x+2 y)$
View full question & answer→Question 393 Marks
Solve:
$ 9 a^2-24 a^2 b^2+16 b^4-256 $
Answer$ 9 a^2-24 a^2 b^2+16 b^4-256 $
$ =\left(9 a^4-24 a^2 b^2+16 b^4\right)-256 $
$ =\left[\left(3 a^2\right)^2-2 \times 3 a^2 \times 4 b^2+\left(4 b^2\right)^2\right]-16^2 $
$ =\left(3 a^2-40^2\right)^2-16^2 $
$ =\left[\left(3 a^2-4 b^2\right)-16\right]\left[\left(3 a^2-4^2\right)+16\right] $
$ =\left(3 a^2-4 b-16\right)\left(3 a^2-4 b^2+16\right) $
View full question & answer→Question 403 Marks
Solve:$(y - 3)(y - 4)$
Answer$(y - 3)(y - 4)= y2 - 7y + 12 $ $\Big[$Adding and subtracting $\Big(\frac{7}{2}\Big)^2\Big]$
$=\text{y}^2-7\text{y}+\Big(\frac{7}{2}\Big)^2-\Big(\frac{7}{2}\Big)^2+12$
Comleting the square
$=\Big(\text{y}-\Big(\frac{7}{2}\Big)\Big)^2-\frac{49}{4}+\frac{48}{4}$
$=\Big(\text{y}-\Big(\frac{7}{2}\Big)\Big)^2-\Big(\frac{1}{4}\Big)$
$=\Big(\text{y}-\Big(\frac{7}{2}\Big)^2-\Big(\frac{1}{2}^2\Big)$
$=\Big[\text{y}-\Big(\frac{7}{2}-\frac{1}{2}\Big)\Big]\Big[\text{y}-\Big(\frac{7}{2}+\frac{1}{2}\Big)\Big]$
$=(\text{y}-4)(\text{y}-3)$
View full question & answer→Question 413 Marks
Factories:
$x^2- 11x - 42$
AnswerTo factories $x^2- 11x - 42,$ we will find two number $p$ and $q$ such that $p + q = -11$ and $pq = -42$
Now,
$3 + (-14) = -22$ And $3 x (-14) = 42$
Splittiong the middle term -11x in the given quadratic as $-14x + 3x$, we get:
$ x^2-11 x-42=x^2-14 x+3 x-42 $
$ =\left(x^2-14\right)+(3 x-42) $
$ =x(x-14)+3(x-14) $
$ =(x-14)(x+3) $
View full question & answer→Question 423 Marks
Solve:
$ x^2-y^2-4 x z+4 z^2 $
Answer$ x^2-y^2-4 x z+4 z^2 $
$ =\left(x^2-4 x z+4 z^2\right)-y^2 $
$ =(x-2 z)^2-y^2 $
$ =[(x-2 z)-y][(x-2 z)+y)] $
$ =(x-2 z-y)(x-2 z+y) $
$ =(x+y-2 z)(x-y-2 z) $
View full question & answer→Question 433 Marks
Find the greatest common factor of the polynomial:
$7 x, 21 x^2$ and $14 x y^2$
AnswerThe numerical coefficients of the given monomials are $7, 21$ and $14.$
The greatest common factor of $7, 21$ and $14$ is $7.$
The common literal appearing in the three monomials is $x.$
The smallest power of $x$ in the three monomials is $1.$
The monomial of the common literals with the smallest powers is $x. $
Hence, the greatest common factor is $7x.$
View full question & answer→Question 443 Marks
Solve:$49(a-b)^2-25(a+b)^2$
Answer
$=\left[7(a-b)^2\right]-[5(a+b)]^2$
$=[7(a-b)-5(a+b)][7(a-b)+5(a+b)] $
$=(7 a-7 b-5 a-5 b)(7 a-7 a+5 a+5 b)$
$=(2 a-12 b)(12-2 b)$
$=2(a-6 b) 2(6 a-b)$
$=4(a-6 b)(6 a-b)$
View full question & answer→Question 453 Marks
Factorize of the following algebraic expressions:$16(2 L-3 m)^2-12(3 m-2 L)$
Answer$16(2 L-3 m)^2-12(3 m-2 L)$
$=16(2 L-3 m)^2+12(2 L-3)[(3 m-2 L)=-(2 L-3 m)]$
$= [16(2L - 3m) + 12] (2L -3)$ [taking $(2L -3)$ as the common factor]
$= 4[4(2L - 3m) + 3] (2L -3)$ [taking 4 as the common factor]
$= 4(8L - 12m +3)(2L - 3m)$
View full question & answer→Question 463 Marks
Solve:$x^2+2 x+1-9 y^2$
Answer$x^2+2 x+1-9 y^2$
$=\left(x^2+2 x+1\right)-9 y^2$
$=\left(x^2+2 \times x \times 1+1\right)-9 y^2$
$=(x+1)^2-(3 y)^2$
$= [(x + 1) - 3Y][(X + 1) - 3y]$
$= (x + 1 - 3y)(x + 1 + 3y)$
$= (x + 3y + 1)(x - 3y + 1)$
View full question & answer→Question 473 Marks
Find the greatest common factor of the polynomial:
$36 a^2 b^2 c^4, 54 a^5 c^2, 90 a^4 b^2 c^2$
AnswerThe numerical coefficients of the given monomials are $36, 54$ and $90.$
The greatest common factor of $36, 54$ and $90$ is $18.$
The common literals appearing in the three monomials are a and $c.$
The smallest power of $a$ in the three monomials is $2.$
The smallest power of $c$ in the three monomials is $2.$
The monomial of common literals with the smallest powers is $a^2c^2$.
Hence, the greatest common factor is $18a^2c^2$.
View full question & answer→Question 483 Marks
Factories:$a^2+14 x+45$
AnswerTo factories $a^2+14 x+45$, we will find two number $p$ and $q$ such that $p+q=14$ and $p q=45$
Now,
$9+5=14 \text { And } 9 \times 5=45$
Splittiong the middle term $14 x$ in the given quadratic as $9 x+5 x$, we get:
$a^2+14 x+45=x^2+9 x+5 x+45$
$=\left(x^2+9 x\right)+(5 x+45)$
$= x(x + 9) + 5(x + 9)$
$= (x + 5)(x + 9)$
View full question & answer→Question 493 Marks
Solve:$49-x^2-y^2+2 x y$
Answer $49-x^2-y^2+2 x y$
$=49-\left(x^2+2 x y-y^2\right)$
$=7^2-(x-y)^2$
$= [7 - (x - y)][7 + (x - y)]$
$= (7 - x + y)(7 + x - y)$
$= (x - y + 7)(y - x + 7)$
View full question & answer→Question 503 Marks
Find the greatest common factor of the polynomial:
$6 x^3 y$ and $18 x^2 y^3$
AnswerThe numerical coefficients of the given monomials are $6$ and $18.$
The greatest common factor of $6$ and $18$ is $6.$
The common literals appearing in the two monomials are x and $y.$
The smallest power of $x$ in the two monomials is $2.$
The smallest power of $y$ in the two monomials is $1.$
The monomial of the common literals with the smallest powers is $x^2y.$
Hence, the greatest common factor is $6x^2y.$
View full question & answer→Question 513 Marks
Solve:$2a^5-32a$
Answer$2 a^5-32 a$
$=2 a\left(a^4-16\right)$
$=2 a\left[\left(a^2\right)^2-4^2\right] $
$=2 a\left(a^2+4\right)\left(a^2-4\right)$
$=2 a\left(a^2+4\right)\left(a^2-2^2\right)$
$=2 a\left(a^2+4\right)(a+2)(a-2)$
$ =2 a(a-a)(a+2)\left(a^2+4\right)$
View full question & answer→Question 523 Marks
Solve:$16-a^6+4 a^3 b^3-4 b^6$
Answer$16-a^6+4 a^3 b^3-4 b^6$
$=16-\left(a^6-4 a^3 b^3+4 b^6\right)$
$ =4^2-\left[\left(a^3\right)^2-2 \times a^3 \times 2 b^3+\left(2 b^3\right)^2\right]$
$ =4^2-\left(a^3-2 b^3\right)^2$
$=\left[4-\left(a^3-2 b^3\right)\right]\left[4+\left(a^3-2 b^3\right)\right]$
$\left.=\left(4-a^3-2 b^3\right)\left(4+a^3-2 b^3\right)\right]$
$=\left(a^3-2 b^3+4\right)\left(-a^3-2 b^3+4\right)$
View full question & answer→Question 533 Marks
Factories:$x^2-22 x+120$
AnswerTo factories $x^2-22 x+120$, we will find two number $p$ and $q$ such that $p+q=-22$ and $p q=120$
Now,
$(-12)+(-10)=-22 \text { And }(-22) \times(-10)=120$
Splittiong the middle term $14 x$ in the given quadratic as $-12 x-10 x$, we get:
$x^2-22 x+12=x^2-12 x-10 x+120$
$=\left(x^2-12\right)+(-10 x+120)$
$= x(x - 12) - 10(x - 12)$
$= (x - 10)(x - 12)$
View full question & answer→Question 543 Marks
Find the greatest common factor of the polynomial:$3 a^2 b^2+4 b^2 c^2+12 a^2 b^2 c^2$
AnswerThe numerical coefficients of the given monomials are $3 a^2 b^2, 4 b^2 c^2$ and $12 a^2 b^2 c^2$.
The greatest common factor of $3 a^2 b^2, 4 b^2 c^2$ and $12 a^2 b^2 c^2$ is $1 .$
The common literal appearing in the three monomials is $b$ .
The smallest power of $b$ in the three monomials is $2 .$
The monomial of common literals with the smallest powers is $b ^2$.
Hence, the greatest common factor is $b^2$.
View full question & answer→Question 553 Marks
Find the greatest common factor of the polynomial: $2 x y z+3 x^2 y+4 y^2$
AnswerThe numerical coefficients of the given monomials are $2 x y z, 3 x^2 y$ and $4 y^2$.
The greatest common factor of $2 x y z, 3 x^2 y$ and $4 y^2$ is $1 .$
The common literal appearing in the three monomials is $y$.
The smallest power of $y$ in the three monomials is $1.$
The monomial of common literals with the smallest powers is $y$.
Hence, the greatest common factor is $y$.
View full question & answer→Question 563 Marks
Solve: $q^2-10 p+21$
Answer$q^2-10 p+21$
$\left.=q^2-10 q+\left(\frac{10}{2}\right)^2-\left(\frac{10}{2}\right)^2+21 \text { [Adding and subtracting }\left(\frac{10}{2}\right)^2, \text { that is } 5^2\right] $
$=q 2-2 \times q \times 5+5^2-5^2+21$
$=(q-5) 2-4[\text { completing the square }]$
$=[(q-5)-2][(q-5)+2]$
$=(q-5-2)(q-5+2)$
$=(q-7)(q-3)$
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