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:
$p^2+6 p-16$
Answer$p^2+6 p-16$
$p^2+6 p+\left(\frac{6}{2}\right)^2-\left(\frac{6}{2}\right)^2-16\left[\text { Adding and suobtrating }\left(\frac{6}{2}\right)^2, \text { that is } 3^2\right]$
$=p^2+6 p+3^2-9-16$
$=(p-3)^2-25[\text { completing the square }]$
$=(p+3)^2-5^2$
$=[(p+3)-5][(p+3)+5]$
$=(p+3-5)(p+3+5)$
$=(p-2)(p+8)$
View full question & answer→Question 53 Marks
Factories:
$a^2-14 a-51$
AnswerTo factories $a^2-14 a-51$, we will find two number $p$ and $q$ such that $p+q=-14$ and $p q=-51$
Now,
$3+(-17)=-14 \text { And } 3 \times(-17)=-51$
Splittiong the middle term -14 in the given quadratic as $3 a-17 a$, we get:
$a^2-14 a-51=a^2+3 a-17 a-51$
$=\left(a^2+3 a\right)-(17 a+51)$
$=a(a+3)-17(a+3)$
$=(a-17)(a+3)$
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 $x^2 y^2$.
Hence, the greatest common factor is $2 x^2 y^2$.
View full question & answer→Question 73 Marks
Solve:
$a^2+4 a b+3 b^2$
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 $15 a$.
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:
$9 x^2, 15 x^2 y^3, 6 x y^2 \text { and } 21 x^2 y^2$
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 $3 x$.
View full question & answer→Question 133 Marks
Factorize:
$2 L^2 m n-3 L m^2 n+4 L m^2$
AnswerThe greatest common factor of the term
$2 L^2 m n-3 L m^2 n$ and $4 L m n^2$ of the expression
$2 L^2 mn -3 Lm ^2 n +4 Lmn ^2$ is Lmn .
Also, we can write $2 L^2 mn = Lmn .2 L, 3 Lm ^2 n = Lmn .4 Lmn ^2= Lmn .4 n$
Therefore, $2 L^2 mn -3 Lm ^2 n +4 Lmn ^2=( Lmn .2 L)-( Lmn .3 m)+( Lmn .4 n )$
$=\operatorname{Lmn}(2 L-3 m+4 n)$
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 $a^2 b^2$.
Hence, the greatest common factor is $a^2 b^2$.
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 $x^2 y z$.
Hence, the greatest common factor is $21 x^2 y z$.
View full question & answer→Question 193 Marks
Find the greatest common factor of the polynomial:
$12 a x^2, 6 a^2 x^3 \text { and } 2 a^3 x^5$
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 $2 a x^2$.
View full question & answer→Question 203 Marks
Factories:
$x^2+12 x-45$
AnswerTo factories $x^2+12 x-45$, we will find two number $p$ and $q$ such that $p+q=12$ and $p q=-45$ Now, $15+(-3)=12$ and $15 \times(-3)=-45$
Splitting the middle term $12 x$ in the given quadratic as $-3 x+15 x$, 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$x^2+9 y^2-6 x y-25 a^2=\left(x^2-6 x y+9 y^2\right)-25 a^2$
$=\left[x^2-2 x x \times 3 y+(3 y)^2\right]-25 a^2$
$=(x-3 y)^2-(5 a) 2$
$=[(x-3 y)-5 a][(x-3 y)+5 a]$
$=(x-3 y-5 a)(x-3 y+5 a)$
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:
$p^2 q-p r^2-p q+r^2$
Answer$p^2 q-p r^2-p q+r^2$
$=\left(p^2 q-p q\right)+\left(r^2-p r^2\right)$
$p q(p-1)+r^2(1-p)$
$p q(p-1)-r^2(p-1)[\text { SINCE, }(1-p)=-(p-1)]$
$=\left(p q-r^2\right)(p-1)[\text { taking }(p-1) \text { as the common factor }]$
View full question & answer→Question 243 Marks
Factorize:
$28 a^2+14 a^2 b^2-21 a^4$
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$25 x^2-10 x+1-36 y^2=\left(25 x^2-10 x+1\right)-36 y^2$
$=\left[(5 x)^2-2 x 5 x x 1+1\right]-36 y^2$
$=(5 x-1)^2-(6 y)^2$
$=[(5 x-1)-6 y][(5 x-1)+6 y]$
$=(5 x-1-6 y)(5 x-1+6 y)$
$=(5 x-6 y-1)(5 x+6 y-1)$
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 $5 a ^4, 10 a ^3$ and $15 a ^2$.
The greatest common factor of $5 a^4, 10 a^3$ and $15 a^2$ 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 $5 a ^2$.
View full question & answer→Question 273 Marks
Solve:
$25-p^2-q^2-2 p q$
Answer$25- p ^2- q ^2-2 pq$
$=25-\left(p^2+2 p q+q^2\right)$
$=5^2-\left(p^2+2 \times p \times q+q^2\right)$
$=5^2-(p+q)^2$
$=[5-(p+q)][5+(p+q)]$
$=(5-p+q)(5+p+q)$
$=-(p+q-5)(p+q+5)$
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:
$(a x+b y)^2+(b x-a y)^2$
Answer$(a x+b y)^2+(b x-a y)^2$
$=a^2 x^2+b^2 y^2+b^2 x^2+a^2 y^2$
$=\left(a^2 x^2+a^2 y^2\right)+\left(b^2 x^2+b^2 y^2\right)$
$=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)$
$=\left(a^2+b^2\right)\left(x^2+y^2\right)\left[\text { taking }\left(x^2+y^2\right)\right. \text { as the common factor] }$
View full question & answer→Question 303 Marks
Factories:
$a^2+3 a-88$
AnswerTo factories $a^2+3 a-88$, we will find two numbers $p$ and $q$ such that $p+q=3$ and $p q=-88$
Now, $11+(-8)=3$
And $11 \times(-8)=-88$
Splitting the middle term $3 a$ in the given quadratic as $11 a-8 a$, we get:
$a^2+3 a-88=a^2+11 a-8 a-88$
$=\left(a^2+11 a\right)-(8 a+88)$
$=a(a+11)-8(a+11)$
$=(a-8)(a+11)$
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 $a^4 b=a b \cdot a^3, 3 a^2 b^2=a b .3 a b$ 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 $x y^2$.
Hence, the greatest common factor is $3 x y^2$.
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 $a^2 b$.
Hence, the greatest common factor is $2 a^2 b$.
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 $x y$.
Hence, the greatest common factor is $2 x y$.
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 $2 x^2$.
View full question & answer→Question 363 Marks
Factories:
$a^2-2 a-3$
AnswerTo factories $a^2-2 a-3$, we will find two number $p$ and $q$ such that $p+q=2$ and $p q=-3$ Now,
$3+(-1)=2 \text { And } 3 \times(-1)=-3$
Splittiong the middle term $2 a$ in the given quadratic as $-a+3 a$, 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:
$10 m^3 n^2+15 m^4 n-20 m^2 n^3$
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$(3 x+4 y)^4-x^4$
$=\left[(3 x+4 y)^2\right]^2-\left(x^2\right)^2$
$\left.\left.=\left[(3 x+4 y)^2+x^2\right][3 x+] 4 y\right)^2-x^2\right]$
$\left.=\left[(3 x+4 y)^2+x^2\right][3 x+4 y)+x\right][(3 x+4 y)-x]$
$=\left\{(3 x+4 y)^2+x^2\right\}(3 x+4 y+x)(3 x+y-x)$
$=\left\{(3 x+4 y)^2+x^2\right\}(4 x+4 y)(2 x+4 y)$
$\left.=\{3 x+4 y)^2+x^2\right\} 4(x+y) 2(x+2 y)$
$=8\left\{(3 x+4 y)^2+x^2\right\}(x+y)(x+2 y)$
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
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: $x^2-11 x-42$
AnswerTo factories $x^2-11 x-42$, we will find two number $p$ and $q$ such that $p+q=-11$ and $p q=-42$ Now,
$3+(-14)=-22$ And $3 \times(-14)=42$
Splittiong the middle term $-11 x$ in the given quadratic as $-14 x+3 x$, 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 7 x .
View full question & answer→Question 443 Marks
Solve:
$49(a-b)^2-25(a+b)^2$
Answer$49(a-b)^2-25(a+b)^2$
$=\left[7(a-b)^2\right]-[5(a+b)]^2$
$=[7(a-b)-5(a+b)][7(a-b)+5(a+b)]$
$=(7 a-7 b-5 a-5 b)(7 a-7 a+5 a+5 b)$
$=(2 a-12 b)(12-2 b)$
$=2(a-6 b) 2(6 a-b)$
$=4(a-6 b)(6 a-b)$
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$16(2 L-3 m)^2-12(3 m-2 L)$
$= 16(2 L-3 m)^2+12(2 L-3)[(3 m-2 L)=-(2 L-3 m)]$
$= {[16(2 L-3 m)+12](2 L-3)[\text { taking }(2 L-3) \text { as the common factor }] }$
$= 4[4(2 L-3 m)+3](2 L-3)[\text { taking } 4 \text { as the common factor] }$
$= 4(8 L-12 m+3)(2 L-3 m)$
View full question & answer→Question 463 Marks
Solve:
$x^2+2 x+1-9 y^2$
Answer$x^2+2 x+1-9 y^2$
$=\left(x^2+2 x+1\right)-9 y^2$
$=\left(x^2+2 \times x \times 1+1\right)-9 y^2$
$=(x+1)^2-(3 y)^2$
$=[(x+1)-3 Y][(x+1)-3 y]$
$=(x+1-3 y)(x+1+3 y)$
$=(x+3 y+1)(x-3 y+1)$
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 $a ^2 c ^2$.
Hence, the greatest common factor is $18 a^2 c^2$.
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 $pq =45$
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 $x^2 y$.
Hence, the greatest common factor is $6 x^2 y$.
View full question & answer→Question 513 Marks
Solve:
$2 a^5-32 a$
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 $pq =120$
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$q^2-10 p+21$
$= q ^2-10 q +\left(\frac{10}{2}\right)^2-\left(\frac{10}{2}\right)^2+21$ [Adding and subtracting $\left(\frac{10}{2}\right)^2$, that is $5^2$ ]
$=q 2-2 \times q \times 5+5^2-5^2+21$
$=(q-5) 2-4$ [ completing the square]
$=[(q-5)-2][(q-5)+2]$
$=(q-5-2)(q-5+2)$
$=(q-7)(q-3)$
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