CHAPTER 8 - ALGEBRAIC EXPRESSIONS AND IDENTITIES

Exercise 8.4

1. Multiply the binomials:
(i) \((2x + 5)\) and \((4x - 3)\)

Using distributive property:

\[ (2x + 5)(4x - 3) = 2x(4x - 3) + 5(4x - 3) \]

\[ = 2x \times 4x + 2x \times (-3) + 5 \times 4x + 5 \times (-3) \]

\[ = 8x^2 - 6x + 20x - 15 \]

Combine like terms:

\[ = 8x^2 + 14x - 15 \]

Final answer: \(8x^2 + 14x - 15\)
(ii) \((2.5l - 0.5m)\) and \((2.5l + 0.5m)\)

This is of the form \((a - b)(a + b) = a^2 - b^2\):

\[ (2.5l - 0.5m)(2.5l + 0.5m)
\]

\[ = (2.5l)^2 - (0.5m)^2 \]

\[ = 6.25l^2 - 0.25m^2 \]

Final answer: \(6.25l^2 - 0.25m^2\)
(v) \((2pq + 3q^2)\) and \((3pq - 2q^2)\)

Multiply:

\[ (2pq + 3q^2)(3pq - 2q^2) \] \[ = 2pq(3pq - 2q^2) + 3q^2(3pq - 2q^2) \]

\[ = 2pq \times 3pq + 2pq \times (-2q^2) \] \[ + 3q^2 \times 3pq + 3q^2 \times (-2q^2) \]

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

Combine like terms:

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

Final answer: \(6p^2q^2 + 5pq^3 - 6q^4\)
(vi) \(\left( \frac{3}{4} a^2 + 3b^2 \right)\) and \(4 \left( a^2 - \frac{2}{3} b^2 \right)\)

First, simplify the second expression:

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

Now multiply:

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

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

\[ = 3a^4 - 2a^2b^2 + 12a^2b^2 - 8b^4 \]

Combine like terms:

\[ = 3a^4 + 10a^2b^2 - 8b^4 \]

Final answer: \(3a^4 + 10a^2b^2 - 8b^4\)
2. Find the product:
(i) \((5 - 2x)(3 + x)\)

Using distributive property:

\[ (5 - 2x)(3 + x) = 5(3 + x) - 2x(3 + x) \]

\[ = 5 \times 3 + 5 \times x - 2x \times 3 - 2x \times x \]

\[ = 15 + 5x - 6x - 2x^2 \]

Combine like terms:

\[ = -2x^2 - x + 15 \]

Final answer: \(-2x^2 - x + 15\)
(ii) \((a^2 + b)(a + b^2)\)

Using distributive property:

\[ (a^2 + b)(a + b^2) = a^2(a + b^2) + b(a + b^2) \]

\[ = a^2 \times a + a^2 \times b^2 + b \times a + b \times b^2 \]

\[ = a^3 + a^2b^2 + ab + b^3 \]

Final answer: \(a^3 + a^2b^2 + ab + b^3\)
(iii) \((x + 7y)(7x - y)\)

Using distributive property:

\[ (x + 7y)(7x - y) = x(7x - y) + 7y(7x - y) \]

\[ = x \times 7x + x \times (-y) + 7y \times 7x + 7y \times (-y) \]

\[ = 7x^2 - xy + 49xy - 7y^2 \]

Combine like terms:

\[ = 7x^2 + 48xy - 7y^2 \]

Final answer: \(7x^2 + 48xy - 7y^2\)
(iv) \((p^2 - q^2)(2p + q)\)

Using distributive property:

\[ (p^2 - q^2)(2p + q) = p^2(2p + q) - q^2(2p + q) \]

\[ = p^2 \times 2p + p^2 \times q - q^2 \times 2p - q^2 \times q \]

\[ = 2p^3 + p^2q - 2pq^2 - q^3 \]

Final answer: \(2p^3 + p^2q - 2pq^2 - q^3\)
3. Simplify:
(i) \((x^2 - 5)(x + 5) + 25\)

First multiply:

\[ (x^2 - 5)(x + 5) = x^2(x + 5) - 5(x + 5) \]

\[ = x^3 + 5x^2 - 5x - 25 \]

Now add 25:

\[ x^3 + 5x^2 - 5x - 25 + 25 = x^3 + 5x^2 - 5x \]

Final answer: \(x^3 + 5x^2 - 5x\)
(ii) \((a^2 + 5)(b^3 + 3) + 5\)

First multiply:

\[ (a^2 + 5)(b^3 + 3) = a^2(b^3 + 3) + 5(b^3 + 3) \]

\[ = a^2b^3 + 3a^2 + 5b^3 + 15 \]

Now add 5:

\[ a^2b^3 + 3a^2 + 5b^3 + 15 + 5\] \[ = a^2b^3 + 3a^2 + 5b^3 + 20 \]

Final answer: \(a^2b^3 + 3a^2 + 5b^3 + 20\)
(iii) \((t + s^2)(t^2 - s)\)

Using distributive property:

\[ (t + s^2)(t^2 - s) = t(t^2 - s) + s^2(t^2 - s) \]

\[ = t \times t^2 + t \times (-s) + s^2 \times t^2 + s^2 \times (-s) \]

\[ = t^3 - ts + t^2s^2 - s^3 \]

Final answer: \(t^3 - ts + t^2s^2 - s^3\)
(iv) \((a + b)(c - d) + (a - b)(c + d) + 2(ac + bd)\)

Expand first each term:

\[ (a + b)(c - d) = a(c - d) + b(c - d)\] \[= ac - ad + bc - bd \]

Expand second term:

\[ (a - b)(c + d) = a(c + d) - b(c + d)\] \[ = ac + ad - bc - bd \]

Expand third term:

\[ 2(ac + bd) = 2ac + 2bd \]

Now add all terms:

\[ (ac - ad + bc - bd) \]\[+ (ac + ad - bc - bd)\] \[ + (2ac + 2bd) \]

\[ = ac + ac + 2ac - ad + ad + bc - bc - bd - bd + 2bd\] \[= 4ac \cancel {-ad} \cancel {+ ad} \cancel {+ bc} \cancel {- bc} \cancel {- 2bd} \cancel {+2 bd} \]

Combine like terms:

\[ = 4ac + 0 + 0 + 0 = 4ac \]

Final answer: \(4ac\)
(v) \((x + y)(2x + y) + (x + 2y)(x - y)\)

First expand \((x + y)(2x + y\)):

\[ (x + y)(2x + y) = x(2x + y) + y(2x + y) \]

\[ = 2x^2 + xy + 2xy + y^2 = 2x^2 + 3xy + y^2 \]

Now expand \((x + 2y)(x - y\)):

\[ (x + 2y)(x - y) = x(x - y) + 2y(x - y) \]

\[ = x^2 - xy + 2xy - 2y^2 = x^2 + xy - 2y^2 \]

Now add both expressions:

\[ (2x^2 + 3xy + y^2) + (x^2 + xy - 2y^2) \]

\[= 2x^2 + x^2 + 3xy + xy + y^2 - 2y^2 \]

\[ = 3x^2 + 4xy - y^2 \]

Final answer: \(3x^2 + 4xy - y^2\)
(vi) \((x + y)(x^2 - xy + y^2)\)

Using distributive property:

\[ (x + y)(x^2 - xy + y^2)\] \[= x(x^2 - xy + y^2) + y(x^2 - xy + y^2) \]

\[ = x \times x^2 + x \times (-xy) + x \times y^2 \] \[+ y \times x^2 + y \times (-xy) + y \times y^2 \]

\[ = x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3 \]

Combine like terms:

\[ = x^3 \cancel {-x^2y} \cancel {+ x^2y} \cancel + {xy^2} \cancel {-xy^2} + y^3 \] \[ = x^3 + y^3 \]

Final answer: \(x^3 + y^3\)
(vii) \((1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y\)

Expand first term:

\[ (1.5x - 4y)(1.5x + 4y + 3) \]

\[= 1.5x(1.5x + 4y + 3)\] \[- 4y(1.5x + 4y + 3) \]

\[ = 1.5x \times 1.5x + 1.5x \times 4y + 1.5x \times 3 \] \[- 4y \times 1.5x - 4y \times 4y - 4y \times 3 \]

\[ = 2.25x^2 + 6xy + 4.5x - 6xy - 16y^2 - 12y \]

Combine like terms:

\[ = 2.25x^2 + (6xy - 6xy) + 4.5x - 16y^2 - 12y \]

\[= 2.25x^2 + 4.5x - 16y^2 - 12y \]

Now subtract \(4.5x - 12y\):

\[ (2.25x^2 + 4.5x - 16y^2 - 12y) - (4.5x - 12y) \]

\[= 2.25x^2 + 4.5x - 16y^2 - 12y - 4.5x + 12y \]

\[ = 2.25x^2 + (4.5x - 4.5x) - 16y^2 + (-12y + 12y) \] \[= 2.25x^2 - 16y^2 \]

Final answer: \(2.25x^2 - 16y^2\)
(viii) \((a + b + c)(a + b - c)\)

Let \(x = a + b\), then:

\[ (x + c)(x - c) = x^2 - c^2 \]

Substitute back:

\[ = (a + b)^2 - c^2 = a^2 + 2ab + b^2 - c^2 \]

Final answer: \(a^2 + 2ab + b^2 - c^2\)

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