3 Marks Question - Maths STD 8 Questions - Vidyadip

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3 Marks Question

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19 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

Factorise the expression and divide them as directed: $39 y^3\left(50 y^2-98\right) \div 26 y^2(5 y+7)$

Answer

$39 y^3\left(50 y^2-98\right) \div 26 y^2(5 y+7)$
$ = \frac{{39{y^3}(50{y^2} - 98)}}{{26{y^2}(5y + 7)}}$
$ = \frac{{39{y^3} \times 2 \times (25{y^2} - 49)}}{{26{y^2}(5y + 7)}}$
$ = \frac{{39{y^3} \times 2 \times \{ {{(5y)}^2} - {{(7)}^2}\} }}{{26{y^2}(5y + 7)}}$
$ = \frac{{39{y^3} \times 2 \times (5y + 7)(5y - 7)}}{{26{y^2}(5y - 7)}}$. . . . [Using Identity III
$= 3y (5y – 7)$

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Question 23 Marks

Factorise the expression and divide them as directed:$ 12 x y\left(9 x^2-16 y^2\right) \div 4 x y(3 x+4 y)$

Answer

$ 12 x y\left(9 x^2-16 y^2\right) \div 4 x y(3 x+4 y)$
$=\frac{{12xy(9{x^2} - 16{y^2})}}{{4xy(3x + 4y)}}$
$ = \frac{{3(9{x^2} - 16{y^2})}}{{3x + 4y}}$
$ = \frac{{3\{ {{(3x)}^2} - (4{y^2})\} }}{{3x + 4y}}$
$ = \frac{{3(3x + 4y)(3x - 4y)}}{{3x + 4y}}$
$= 3(3x – 4y)$

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Question 33 Marks

Factorise the expression and divide them as directed: $5pq (p^2– q^2)$ $\div$ $2p (p + q)$

Answer

$5pq (p^2– q^2)$ $\div$ $2p (p + q)$
$ = \frac{{5pq({p^2} - {q^2})}}{{2p(p + q)}}$
$ = \frac{{5pq(p + q)(p - q)}}{{2p(p + q)}}$. . . . [Using Identity $III]$
$ = {5q(p-q) \over 2}$

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Question 43 Marks

Factorise the expression and divide them as directed: $4yz (z^2+ 6z – 16) \div 2y (z + 8)$

Answer

$4yz (z^2+ 6z – 16) \div 2y (z + 8)$
$ = \frac{{4yz({z^2} + 6z - 16)}}{{2y(z + 8)}}$
$ = \frac{{2z({z^2} + 6z - 16)}}{{z + 8}}$
$ = \frac{{2z({z^2} + 8z - 2z - 16)}}{{z + 8}}$. . . . [Using Identity $IV]$
$ = \frac{{2z[z(z + 8) - 2(z + 8)]}}{{z + 8}}$
$ = \frac{{2z(z + 8)(z - 2)}}{{z + 8}}$
$= 2z (z – 2)$

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Question 53 Marks

Factorise the expression and divide them as directed: $(5p^2– 25p + 20) \div (p – 1)$

Answer

$(5p^2– 25p + 20) \div (p – 1)$
$ = \frac{{5({p^2} - 5p + 4)}}{{p - 1}}$
$ = \frac{{5({p^2} - p - 4p + 4)}}{{p - 1}}$. . . . [Applying Identity $IV]$
$ = \frac{{5\{ p(p - 1) - 4(p - 1)\} }}{{p - 1}}$
$ = \frac{{5(p - 1)(p - 4)}}{{p - 1}}$
$= 5 (p – 4)$

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Question 63 Marks

Factorise the expression and divide them as directed: $(m^2– 14m – 32)\div (m + 2)$

Answer

$(m^2– 14m – 32) \div(m + 2)$
$ = \frac{{{m^2} - 14m - 32}}{{m + 2}}$
$ = \frac{{{m^2} - 16m + 2m - 32}}{{m + 2}}$. . . . [Using Identity $IV$
$ = \frac{{m(m - 16) + 2(m - 16)}}{{m + 2}}$
$ = \frac{{m(m - 16)(m + 2)}}{{m + 2}}$
$= m – 16$

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Question 73 Marks

Factorise the expression and divide them as directed: $\left(y^2+7 y+10\right) \div(y+5)$

Answer

$\left(y^2+7 y+10\right) \div(y+5)$
$ = \frac{{{y^2} + 7y + 10}}{{y + 5}}$
$ = \frac{{{y^2} + 2y + 5y + 10}}{{y + 5}}$ . . . .[Using Identity $IV$
$ = \frac{{y(y + 2) + 5(y + 2)}}{{y + 5}}$
$ = \frac{{(y + 2)(y + 5)}}{{y + 5}}$
$= y + 2$

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Question 83 Marks

Factorise the expressions: $p^2 6p – 16$

Answer

$p^2+6 p-16$
$=p^2+6 p+9-25$
$=(p)^2+2(p)(3)+(3)^2-(5)^2$
$=(p+3)^2-(5)^2. . . . $[Using Identity $I$
$= (p + 3 + – 5) (p + 3 + 5) . . . .$ [Applying Identity $III]$
$= (p – 2) (p + 8).$

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Question 93 Marks

Factorise the expressions: $q^2- 10q + 21$

Answer

$q^2– 10q + 21$
$= q^2– 10q + 25 – 4$
$= {(q)^2– 2(q) (5) + (5)^2} – 4$
$= (q – 5)^2– (2)^2. . . .$ [Using Identity $II]$
$= (q – 5 – 2) (q – 5 + 2). . . .$ [Using Identity $III]$
$= (q – 7) (q – 3)$

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Question 103 Marks

Factorise the expressions: $p^2+ 6p + 8$

Answer

$ p^2+6 p+8 $
$ =p^2+6 p+9-1 $
$ =\left\{(p)^2+2(p)(3)+(3)^2\right\}-(1)^2 $
$ =(p+3)^2-(1)^2 . . . .$ [Using Identity $I$
$= (p + 3 – 1) (p + 3 + 1). . . .$ [Using Identity $III]$
$= (p + 2) (p + 4)$

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Question 113 Marks

Factorise: $a^4-2 a^2 b^2+b^4$

Answer

$a^4-2 a^2 b^2+b^4$
$a^4-2 a^2 b^2+b^4$ $=\left(a^2\right)^2-2\left(a^2\right)\left(b^2\right)+\left(b^2\right)^2$
$ =\left(a^2-b^2\right)^2. . . .$ [Using Identity $II$
$=\left\{(a-b)(a+b)^2\right\}. . . .$ [Using Identity $III$
$=(a-b)^2(a+b)^2$.

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Question 123 Marks

Factorise: $x^4– (x – z)^4$

Answer

$x^4-(x-z)^4$
$\left(x^2\right)^2-\left\{(x-z)^2\right\}^2. . . .$ [Using Identity $III$
$=\left\{x^2-(x-z)^2\right\}\left\{x^2+(x-z)^2\right\} . . . .$ [Applying Identity $III$
$=(x-x+z)(x+x-z)\left\{x^2+(x-z)^2\right\}$
$=z(2 x-z)\left\{x^2+(x-z)^2\right\}$
$=z(2 x-z)\left(x^2+x^2-2 x z+z^2\right). . . . $[Using Identity $II$
$=z(2 x-z)\left(2 x^2-2 x z+z^2\right)$

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Question 133 Marks

Factorise: $25a^2– 4b^2+ 28bc – 49c^2$

Answer

$25a^2– 4b^2+ 28bc – 49c^2$
$= 25a^2– (4b^2– 28bc + 49c^2)$
$= 25a^2– {(2b)^2– 2(2b) (7c) + (7c)^2}$
$= (5a)^2– (2b – 7c)^2. . . .$ [Using Identity $II$
$= {5a – (2b – 7c)} {5a + (2b – 7c)}$
$= (5a – 2b + 7c) (5a + 2b – 7c)$

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Question 143 Marks

Factorise the expression: $(l + m)^2– 4lm$
(Hint: Expand $( l + m)^2$ first)

Answer

$(l + m)^2– 4lm$
$= (l^2+ 2lm + m^2) – 4lm. . . .$ [Using Identity $I]$
$= l^2+ (2lm – 4lm) + m^2 . . .$ [Combining the like terms]
$= l^2– 2lm + m^2$
$= (l)^2– 2(l) (m) + (m)^2$
$= (l – m)^2 . . .$ [Applying Identity $II]$

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Question 153 Marks

Factorise: $m^4- 256$

Answer

We know that, $\mathrm{m}^4=\left(\mathrm{m}^2\right)^2$
and $256=(16)^2$
Therefore, $\mathrm{m}^4-256=\left(\mathrm{m}^2\right)^2-(16)^2$
$=\left(m^2+16\right)\left(m^2-16\right)[$using identity $\left.a^2-b^2=(a+b)(a-b)\right]$
$=\left(m^2+16\right)\left(m^2-4^2\right)$
$=\left(m^2+16\right)(m+4)(m-4)[$again, using identity $\left.a^2-b^2=(a+b)(a-b)\right]$

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Question 163 Marks

Factorise: $10x^2– 18x^3+ 14x^4$

Answer

$10x^2= 2 \times 5 \times x \times x$
$ 18x^3= 2 \times 3 \times 3 \times x \times x \times x$
$ 14x^4= 2 \times 7 \times x \times x \times x \times x$
The common factors of the three terms are $2, x$ and $x.$
Therefore, $10x^2– 18x^3+ 14x^4= (2 \times x \times x \times 5) – (2 \times x \times x \times 3 \times 3 \times x) + (2 \times x \times x \times 7 \times x \times x)$
$= 2 \times x \times x \times [5 – (3 \times 3 \times x) + (7 \times x \times x)]$ (combining the three terms)
$= 2x^2(5 – 9x + 7x^2) = 2x^2(7x^2- 9x + 5)$

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Question 173 Marks

Divide: $z(5z^2- 80)$ by $5z(z + 4)$

Answer

$z(5z^2– 80)$
$= z[(5 \times z^2) – (5 \times 16)]$
$= z \times 5 \times (z^2– 16)$
$= 5z(z + 4)(z – 4)$ [using the identity $a^2– b^2= (a + b) (a – b)]$
Thus, $z(5z^2– 80) ÷ 5z(z + 4)$
$= \frac{5 z(z-4)(z+4)}{5 z(z+4)} = (z – 4)$

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Question 183 Marks

Find the factors of $3m^2+ 9m + 6.$

Answer

$ 3 m^2+9 m+6 $
$ =3\left(m^2+3 m+2\right) $
$ =3\left(m^2+m+2 m+2\right) $
$ =3[m(m+1)+2(m+1)] $
$ =3(m+1)(m+2) $
Therefore, the factors of $3m^2+ 9m + 6$ are $3, (m + 1)$ and $(m + 2).$

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Question 193 Marks

Factorise: $12a^2b + 15ab^2$

Answer

We have $12a^2b = 2 \times 2 \times 3 \times a \times a \times b $
$ 15ab^2= 3 \times 5 \times a \times b \times b$
The two terms have $3, a$ and $b$ as common factors.
Therefore, $12a^2b + 15ab^2= (3 \times a \times b \times 2 \times 2 \times a) + (3 \times a \times b \times 5 \times b)$
$= 3 \times a \times b \times [(2 \times 2 \times a) + (5 \times b)]$ (combining the terms)
$= 3ab \times (4a + 5b)$
$= 3ab (4a + 5b)$ (required factor form)

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