Ncert solutions

Ncert solutions

Grade 8

Algebraic Expressions and Identities | Exercise 9.4

Question 1

Q1) Multiply the binomials:

(i) \left(2x+5\right)and\left(4x-3\right)

(ii) \left(y-8\right)\ and\ \left(3y-4\right)

(iii) \left(2.5l-0.5m\right)\ and\ \left(2.5l+0.5m\right)

(iv) \left(a+3b\right)\ and\ \ \left(x+5\right)

(v) \left(2pq+3q^2\right)\ and\ \left(3pq-2q^2\right)

(vi) \left(\frac{3}{4}a^2+3b^2\right)\ and\ 4\left(a^2-\frac{2}{3}b^2\right)

Solution:

(i) \left(2x+5\right)\times\left(4x-3\right)

= 2x\left(4x-3\right)+5\left(4x-3\right)

= 2x\times4x-2x\times3+5\times4x-5\times3

= 8x^2-6x+20x-15

= 8x^2+14x-15

(ii) \left(y-8\right)\times\left(3y-4\right)=y\left(3y-4\right)-8\left(3y-4\right)

= y\times3y-y\times4-8\times3y-8\times-4

= 3y^2-4y-24y+12

= 3y^2-28y+12

(iii) \left(2.5l\ -\ 0.5m\right)\times\left(2.5l+0.5m\right)

= 2.5l\times\left(2.5l+0.5m\right)-0.5m\times\left(2.5l+0.5m\right)

= 2.5l\times2.5l+0.5l\times0.5m-0.5m\times2.5l-0.5m\times0.5m

= 6.25l^2+1.25lm-1.25lm-0.25m^2

= 6.25l^2-0.25m^2

(iv) \left(a+3b\right)\times\left(x+5\right)=a\left(x+5\right)+3b\left(x+5\right)

= a\times x+a\times5+3b\times x+3b\times5

= ax + 5a + 3bx + 15b

(v) \left(2pq+3q^2\right)\left(3pq-2q^2\right)

= 2pq\times\left(3pq-2q^2\right)+3q^2\left(3pq-2q^2\right)

= 2pq\times3pq-2pq\times2q^2+3q^2\times3pq-3q^2\times2q^2

= 6p^2q^2-4pq^3+9pq^2-6q^4

= 6p^2q^2+5pq^3-6q^4

(vi) \left(\frac{3}{4}a^2+3b^2\right)\times4\left(a^2-\frac{2}{3}b^2\right)

= \left(\frac{3}{4}a^2+3b^2\right)\times\left(4a^2-\frac{8}{3}b^2\right)

= \frac{3}{4}a^2\times\left(4a^2-\frac{8}{3}b^2\right)+3b^2\times\left(4a^2-\frac{8}{3}b^2\right)

= \frac{3}{4}a^2\times4a^2-\frac{3}{4}a^2\times\frac{8}{3}b^2+3b^2\times4a^2-3b^2\times\frac{8}{3}b^2

= 3a^4-2a^2b^2+12a^2b^2-8b^4

= 3a^4+10a^2b^2-8b^4

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subject-cta

Question 1

Q1) Multiply the binomials:

(i) \left(2x+5\right)and\left(4x-3\right)

(ii) \left(y-8\right)\ and\ \left(3y-4\right)

(iii) \left(2.5l-0.5m\right)\ and\ \left(2.5l+0.5m\right)

(iv) \left(a+3b\right)\ and\ \ \left(x+5\right)

(v) \left(2pq+3q^2\right)\ and\ \left(3pq-2q^2\right)

(vi) \left(\frac{3}{4}a^2+3b^2\right)\ and\ 4\left(a^2-\frac{2}{3}b^2\right)

Solution:

(i) \left(2x+5\right)\times\left(4x-3\right)

= 2x\left(4x-3\right)+5\left(4x-3\right)

= 2x\times4x-2x\times3+5\times4x-5\times3

= 8x^2-6x+20x-15

= 8x^2+14x-15

(ii) \left(y-8\right)\times\left(3y-4\right)=y\left(3y-4\right)-8\left(3y-4\right)

= y\times3y-y\times4-8\times3y-8\times-4

= 3y^2-4y-24y+12

= 3y^2-28y+12

(iii) \left(2.5l\ -\ 0.5m\right)\times\left(2.5l+0.5m\right)

= 2.5l\times\left(2.5l+0.5m\right)-0.5m\times\left(2.5l+0.5m\right)

= 2.5l\times2.5l+0.5l\times0.5m-0.5m\times2.5l-0.5m\times0.5m

= 6.25l^2+1.25lm-1.25lm-0.25m^2

= 6.25l^2-0.25m^2

(iv) \left(a+3b\right)\times\left(x+5\right)=a\left(x+5\right)+3b\left(x+5\right)

= a\times x+a\times5+3b\times x+3b\times5

= ax + 5a + 3bx + 15b

(v) \left(2pq+3q^2\right)\left(3pq-2q^2\right)

= 2pq\times\left(3pq-2q^2\right)+3q^2\left(3pq-2q^2\right)

= 2pq\times3pq-2pq\times2q^2+3q^2\times3pq-3q^2\times2q^2

= 6p^2q^2-4pq^3+9pq^2-6q^4

= 6p^2q^2+5pq^3-6q^4

(vi) \left(\frac{3}{4}a^2+3b^2\right)\times4\left(a^2-\frac{2}{3}b^2\right)

= \left(\frac{3}{4}a^2+3b^2\right)\times\left(4a^2-\frac{8}{3}b^2\right)

= \frac{3}{4}a^2\times\left(4a^2-\frac{8}{3}b^2\right)+3b^2\times\left(4a^2-\frac{8}{3}b^2\right)

= \frac{3}{4}a^2\times4a^2-\frac{3}{4}a^2\times\frac{8}{3}b^2+3b^2\times4a^2-3b^2\times\frac{8}{3}b^2

= 3a^4-2a^2b^2+12a^2b^2-8b^4

= 3a^4+10a^2b^2-8b^4

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