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Chapter 10

Algebraic Expressions and Identities — Exercise 10.4

Class - 8 ML Aggarwal Understanding ICSE Mathematics

Exercise 10.4

Question 1(i)

Divide:

−39pq2r5-39pq^2r^5 by −24p3q3r-24p^3q^3r

Answer

Solving,

⇒−39pq2r5÷(−24p3q3r)⇒−39pq2r5−24p3q3r⇒3924×pq2r5p3q3r⇒138×r5−1p3−1q3−2⇒13r48p2q\Rightarrow -39pq^2r^5 \div (-24p^3q^3r) \\[1em] \Rightarrow \dfrac{-39pq^2r^5}{-24p^3q^3r}\\[1em] \Rightarrow \dfrac{39}{24} \times \dfrac{pq^2r^5}{p^3q^3r}\\[1em] \Rightarrow \dfrac{13}{8} \times \dfrac{r^{5-1}}{p^{3-1}q^{3-2}}\\[1em] \Rightarrow \dfrac{13r^4}{8p^2q}

Hence, the required quotient = 13r48p2q\bm{\dfrac{13r^4}{8p^2q}}.

Question 1(ii)

Divide:

−34a2b3-\dfrac{3}{4}a^2b^3 by 67a3b2\dfrac{6}{7}a^3b^2

Answer

Solving,

⇒−34a2b3÷67a3b2⇒−34a2b3×76a3b2⇒−3×74×6×a2b3a3b2⇒−2124×b3−2a3−2⇒−7b8a\Rightarrow -\dfrac{3}{4}a^2b^3 \div \dfrac{6}{7}a^3b^2\\[1em] \Rightarrow -\dfrac{3}{4}a^2b^3 \times \dfrac{7}{6a^3b^2}\\[1em] \Rightarrow -\dfrac{3 \times 7}{4 \times 6} \times \dfrac{a^2b^3}{a^3b^2}\\[1em] \Rightarrow -\dfrac{21}{24} \times \dfrac{b^{3-2}}{a^{3-2}}\\[1em] \Rightarrow -\dfrac{7b}{8a}

Hence, the required quotient = −7b8a\bm{-\dfrac{7b}{8a}}.

Question 2(i)

Divide:

9x4−8x3−12x+39x^4 - 8x^3 - 12x + 3 by 3x3x

Answer

Solving,

⇒9x4−8x3−12x+33x⇒9x43x−8x33x−12x3x+33x⇒3x4−1−83x3−1−4+1x⇒3x3−83x2−4+1x\Rightarrow \dfrac{9x^4 - 8x^3 - 12x + 3}{3x}\\[1em] \Rightarrow \dfrac{9x^4}{3x} - \dfrac{8x^3}{3x} - \dfrac{12x}{3x} + \dfrac{3}{3x}\\[1em] \Rightarrow 3x^{4-1} - \dfrac{8}{3}x^{3-1} - 4 + \dfrac{1}{x}\\[1em] \Rightarrow 3x^3 - \dfrac{8}{3}x^2 - 4 + \dfrac{1}{x}

Hence, the required quotient = 3x3−83x2−4+1x\bm{3x^3 - \dfrac{8}{3}x^2 - 4 + \dfrac{1}{x}}.

Question 2(ii)

Divide:

14p2q3−32p3q2+15pq2−22p+18q14p^2q^3 - 32p^3q^2 + 15pq^2 - 22p + 18q by −2p2q-2p^2q

Answer

Solving,

⇒14p2q3−32p3q2+15pq2−22p+18q−2p2q⇒14p2q3−2p2q−32p3q2−2p2q+15pq2−2p2q−22p−2p2q+18q−2p2q⇒−7q3−1+16p3−2q2−1−15q2−12p2−1+11pq−9p2⇒−7q2+16pq−15q2p+11pq−9p2\Rightarrow \dfrac{14p^2q^3 - 32p^3q^2 + 15pq^2 - 22p + 18q}{-2p^2q}\\[1em] \Rightarrow \dfrac{14p^2q^3}{-2p^2q} - \dfrac{32p^3q^2}{-2p^2q} + \dfrac{15pq^2}{-2p^2q} - \dfrac{22p}{-2p^2q} + \dfrac{18q}{-2p^2q}\\[1em] \Rightarrow -7q^{3-1} + 16p^{3-2}q^{2-1} - \dfrac{15q^{2-1}}{2p^{2-1}} + \dfrac{11}{pq} - \dfrac{9}{p^2}\\[1em] \Rightarrow -7q^2 + 16pq - \dfrac{15q}{2p} + \dfrac{11}{pq} - \dfrac{9}{p^2}

Hence, the required quotient = −7q2+16pq−15q2p+11pq−9p2\bm{-7q^2 + 16pq - \dfrac{15q}{2p} + \dfrac{11}{pq} - \dfrac{9}{p^2}}.

Question 3(i)

Divide:

6x2+13x+56x^2 + 13x + 5 by 2x+12x + 1

Answer

Dividing 6x2+13x+56x^2 + 13x + 5 by 2x+12x + 1

2x+1)3x+52x+1)6x2+13x+5‾+++−6x2+−3x+5‾++++++10x+5+++++()−10x+−5‾2x+1)6x2+13x;0\begin{array}{l} \phantom{2x + 1\big)}{3x + 5} \\ 2x + 1\overline{\smash{\big)}6x^2 + 13x + 5} \\ \phantom{+++}\underline{\underset{-}{}6x^2 \underset{-}{+} 3x \phantom{+ 5}} \\ \phantom{++++++}10x + 5 \\ \phantom{+++++()}\underline{\underset{-}{}10x \underset{-}{+} 5} \\ \phantom{2x + 1\big)6x^2 + 13x;}0 \end{array}

Hence, quotient = 3x+5\bm{3x + 5} and remainder = 0.

Question 3(ii)

Divide:

1+y31 + y^3 by 1+y1 + y

Answer

Arranging the terms of the dividend and the divisor in descending powers of yy, we divide y3+1y^3 + 1 by y+1y + 1

y+1)y2−y+1y+1)y3+1‾++−y3+−y2+1‾y+1)y3;−y2+1y+1)y3;−+y2;−+y+1‾y+1)y3+1;y+1y+1)y3+1;−y+−1‾+++++++0\begin{array}{l} \phantom{y + 1\big)}{y^2 - y + 1} \\ y + 1\overline{\smash{\big)}y^3 + 1} \\ \phantom{++}\underline{\underset{-}{}y^3 + \underset{-}{}y^2 \phantom{+ 1}} \\ \phantom{y + 1\big)y^3;}-y^2 + 1 \\ \phantom{y + 1\big)y^3;}\underline{\underset{+}{-}y^2 ; -\underset{+}{}y \phantom{+ 1}} \\ \phantom{y + 1\big)y^3 + 1;}y + 1 \\ \phantom{y + 1\big)y^3 + 1;}\underline{\underset{-}{}y \underset{-}{+} 1} \\ \phantom{+++++++}0 \end{array}

Hence, quotient = y2−y+1\bm{y^2 - y + 1} and remainder = 0.

Question 3(iii)

Divide:

5+x−2x25 + x - 2x^2 by x+1x + 1

Answer

Arranging the terms of the dividend in descending powers of xx, we divide −2x2+x+5-2x^2 + x + 5 by x+1x + 1

x+1)−2x+3x+1)−2x2+x+5‾x+1)−+2x2−+2x+5‾++++++3x+5x+1)−2x2;−3x+−3‾x+1)−2x2+x;2\begin{array}{l} \phantom{x + 1\big)}{-2x + 3} \\ x + 1\overline{\smash{\big)}-2x^2 + x + 5} \\ \phantom{x + 1\big)}\underline{\underset{+}{-}2x^2 \underset{+}{-} 2x \phantom{+ 5}} \\ \phantom{++++++}3x + 5 \\ \phantom{x + 1\big)-2x^2;}\underline{\underset{-}{}3x \underset{-}{+} 3} \\ \phantom{x + 1\big)-2x^2 + x;}2 \end{array}

Hence, quotient = −2x+3\bm{-2x + 3} and remainder = 2.

Question 3(iv)

Divide:

x3−6x2+12x−8x^3 - 6x^2 + 12x - 8 by x−2x - 2

Answer

Dividing x3−6x2+12x−8x^3 - 6x^2 + 12x - 8 by x−2x - 2

x−2)x2−4x+4x−2)x3−6x2+12x−8‾++−x3−+2x2+12x−8‾++++−4x2+12x−8x−2)x3;−+4x2+−8x−8‾++++++++4x−8+++++++()−4x−+8‾x−2)x3−6x2+12x;0\begin{array}{l} \phantom{x - 2\big)}{x^2 - 4x + 4} \\ x - 2\overline{\smash{\big)}x^3 - 6x^2 + 12x - 8} \\ \phantom{++}\underline{\underset{-}{}x^3 \underset{+}{-} 2x^2 \phantom{+ 12x} \phantom{- 8}} \\ \phantom{++++}-4x^2 + 12x - 8 \\ \phantom{x - 2\big)x^3;}\underline{\underset{+}{-}4x^2 \underset{-}{+} 8x \phantom{- 8}} \\ \phantom{++++++++}4x - 8 \\ \phantom{+++++++()}\underline{\underset{-}{}4x \underset{+}{-} 8} \\ \phantom{x - 2\big)x^3 - 6x^2 + 12x;}0 \end{array}

Hence, quotient = x2−4x+4\bm{x^2 - 4x + 4} and remainder = 0.

Question 4(i)

Divide:

6x3+x2−26x−256x^3 + x^2 - 26x - 25 by 3x−73x - 7

Answer

Dividing 6x3+x2−26x−256x^3 + x^2 - 26x - 25 by 3x−73x - 7

3x−7)2x2+5x+33x−7)6x3+x2−26x−25‾+++−6x3−+14x2−26x−25‾++++++15x2−26x−25+++++()−15x2−+35x−25‾++++++++++9x−25+++++++++()−9x−+21‾++++++++++++−4\begin{array}{l} \phantom{3x - 7\big)}{2x^2 + 5x + 3} \\ 3x - 7\overline{\smash{\big)}6x^3 + x^2 - 26x - 25} \\ \phantom{+++}\underline{\underset{-}{}6x^3 \underset{+}{-} 14x^2 \phantom{- 26x} \phantom{- 25}} \\ \phantom{++++++}15x^2 - 26x - 25 \\ \phantom{+++++()}\underline{\underset{-}{}15x^2 \underset{+}{-} 35x \phantom{- 25}} \\ \phantom{++++++++++}9x - 25 \\ \phantom{+++++++++()}\underline{\underset{-}{}9x \underset{+}{-} 21} \\ \phantom{++++++++++++}-4 \end{array}

Hence, quotient = 2x2+5x+3\bm{2x^2 + 5x + 3} and remainder = -4.

Question 4(ii)

Divide:

m3−6m2+7m^3 - 6m^2 + 7 by m−1m - 1

Answer

Dividing m3−6m2+7m^3 - 6m^2 + 7 by m−1m - 1

m−1)m2−5m−5m−1)m3−6m2+7‾++()−m3−+m2+7‾m−1)m3;−5m2+7+++++−+5m2+−5m+7‾m−1)m3−6m2;−5m+7++++++++−+5m+−5‾+++++++++++2\begin{array}{l} \phantom{m - 1\big)}{m^2 - 5m - 5} \\ m - 1\overline{\smash{\big)}m^3 - 6m^2 + 7} \\ \phantom{++()}\underline{\underset{-}{}m^3 \underset{+}{-} m^2 \phantom{+ 7}} \\ \phantom{m - 1\big)m^3;}-5m^2 + 7 \\ \phantom{+++++}\underline{\underset{+}{-}5m^2 + \underset{-}{}5m \phantom{+ 7}} \\ \phantom{m - 1\big)m^3 - 6m^2;}-5m + 7 \\ \phantom{++++++++}\underline{\underset{+}{-}5m \underset{-}{+} 5} \\ \phantom{+++++++++++}2 \end{array}

Hence, quotient = m2−5m−5\bm{m^2 - 5m - 5} and remainder = 2.

Question 5(i)

Divide:

a3+2a2+2a+1a^3 + 2a^2 + 2a + 1 by a2+a+1a^2 + a + 1

Answer

Dividing a3+2a2+2a+1a^3 + 2a^2 + 2a + 1 by a2+a+1a^2 + a + 1

a2+a+1)a+1a2+a+1)a3+2a2+2a+1‾a2+a+1)−a3+−a2+−a+1‾++++++++a2+a+1++++++++−a2+−a+−1‾a2+a+1)a3+2a2+2a;0\begin{array}{l} \phantom{a^2 + a + 1\big)}{a + 1} \\ a^2 + a + 1\overline{\smash{\big)}a^3 + 2a^2 + 2a + 1} \\ \phantom{a^2 + a + 1\big)}\underline{\underset{-}{}a^3 \underset{-}{+} a^2 \underset{-}{+} a \phantom{+ 1}} \\ \phantom{++++++++}a^2 + a + 1 \\ \phantom{++++++++}\underline{\underset{-}{}a^2 \underset{-}{+} a \underset{-}{+} 1} \\ \phantom{a^2 + a + 1\big)a^3 + 2a^2 + 2a;}0 \end{array}

Hence, quotient = a+1\bm{a + 1} and remainder = 0.

Question 5(ii)

Divide:

12x3−17x2+26x−1812x^3 - 17x^2 + 26x - 18 by 3x2−2x+53x^2 - 2x + 5

Answer

Dividing 12x3−17x2+26x−1812x^3 - 17x^2 + 26x - 18 by 3x2−2x+53x^2 - 2x + 5

3x2−2x+5)4x−33x2−2x+5)12x3−17x2+26x−18‾3x2−2x+5)−12x3−+8x2+−20x−18‾+++++++++−9x2+6x−18+++++++++()−+9x2+−6x−+15‾3x2−2x+5)12x3−17x2+26x;−3\begin{array}{l} \phantom{3x^2 - 2x + 5\big)}{4x - 3} \\ 3x^2 - 2x + 5\overline{\smash{\big)}12x^3 - 17x^2 + 26x - 18} \\ \phantom{3x^2 - 2x + 5\big)}\underline{\underset{-}{}12x^3 \underset{+}{-} 8x^2 \underset{-}{+} 20x \phantom{- 18}} \\ \phantom{+++++++++}-9x^2 + 6x - 18 \\ \phantom{+++++++++()}\underline{\underset{+}{-}9x^2 \underset{-}{+} 6x \underset{+}{-} 15} \\ \phantom{3x^2 - 2x + 5\big)12x^3 - 17x^2 + 26x;}-3 \end{array}

Hence, quotient = 4x−3\bm{4x - 3} and remainder = -3.

Question 6

If the area of a rectangle is 8x2−45y2+18xy8x^2 - 45y^2 + 18xy and one of its sides is 4x+15y4x + 15y, find the length of adjacent side.

Answer

Area of the rectangle = 8x2−45y2+18xy8x^2 - 45y^2 + 18xy

One side = 4x+15y4x + 15y

As we know, Area of rectangle = One side × Adjacent side

∴ Adjacent side = Area of rectangle ÷ One side

Arranging the terms of the dividend in descending powers of xx, we divide 8x2+18xy−45y28x^2 + 18xy - 45y^2 by 4x+15y4x + 15y

4x+15y)2x−3y4x+15y)8x2+18xy−45y2‾4x+15y)−8x2+−30xy−45y2‾4x+15y)8x2;−12xy−45y24x+15y)8x2;−+12xy−+45y2‾4x+15y)8x2+18xy;0\begin{array}{l} \phantom{4x + 15y\big)}{2x - 3y} \\ 4x + 15y\overline{\smash{\big)}8x^2 + 18xy - 45y^2} \\ \phantom{4x + 15y\big)}\underline{\underset{-}{}8x^2 \underset{-}{+} 30xy \phantom{- 45y^2}} \\ \phantom{4x + 15y\big)8x^2;}-12xy - 45y^2 \\ \phantom{4x + 15y\big)8x^2;}\underline{\underset{+}{-}12xy \underset{+}{-} 45y^2} \\ \phantom{4x + 15y\big)8x^2 + 18xy;}0 \end{array}

Hence, the length of the adjacent side = 2x−3y\bm{2x - 3y}.

Question 7

What should be subtracted from the polynomial x4+2x3+x2−18x−15x^4 + 2x^3 + x^2 - 18x - 15 so that the resulting polynomial is exactly divisible by x2−1−xx^2 - 1 - x

Answer

If a polynomial is divided by another polynomial, then on subtracting the remainder from the dividend, the resulting polynomial becomes exactly divisible by the divisor.

So, we first find the remainder on dividing x4+2x3+x2−18x−15x^4 + 2x^3 + x^2 - 18x - 15 by x2−1−xx^2 - 1 - x.

Arranging the terms of the divisor in descending powers of xx, we divide x4+2x3+x2−18x−15x^4 + 2x^3 + x^2 - 18x - 15 by x2−x−1x^2 - x - 1

x2−x−1)x2+3x+5x2−x−1)x4+2x3+x2−18x−15‾x2−x−1)−x4−+x3−+x2−18x−15‾+++++++)3x3+2x2−18x−15+++++++−3x3−+3x2−+3x−15‾++++++++++5x2−15x−15+++++++++()−5x2−+5x−+5‾++++++++++++−10x−10\begin{array}{l} \phantom{x^2 - x - 1\big)}{x^2 + 3x + 5} \\ x^2 - x - 1\overline{\smash{\big)}x^4 + 2x^3 + x^2 - 18x - 15} \\ \phantom{x^2 - x - 1\big)}\underline{\underset{-}{}x^4 \underset{+}{-} x^3 \underset{+}{-} x^2 \phantom{- 18x} \phantom{- 15}} \\ \phantom{+++++++)}3x^3 + 2x^2 - 18x - 15 \\ \phantom{+++++++}\underline{\underset{-}{}3x^3 \underset{+}{-} 3x^2 \underset{+}{-} 3x \phantom{- 15}} \\ \phantom{++++++++++}5x^2 - 15x - 15 \\ \phantom{+++++++++()}\underline{\underset{-}{}5x^2 \underset{+}{-} 5x \underset{+}{-} 5} \\ \phantom{++++++++++++}-10x - 10 \end{array}

Here, quotient = x2+3x+5x^2 + 3x + 5 and remainder = −10x−10-10x - 10

Hence, −10x−10\bm{-10x - 10} should be subtracted from the given polynomial.

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