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# 2.Use the identity $$(x + a)(x + b) = x^2 + (a + b)x + ab$$ to find the following products. (i) $$(x + 3) (x + 7)$$ (ii) $$(4x + 5)(4x + 1)$$ (iii)$$(4x – 5) (4x – 1)$$ (iv) $$(4x + 5) (4x – 1)$$ (v) $$(2x + 5y) (2x + 3y)$$ (vi) $$(2a2 + 9) (2a2 + 5)$$ (vii) $$(xyz – 4) (xyz – 2)$$

(i)We have: $$(x + 3) (x + 7)$$

$$=x^2 + (3 + 7)x + 3\times 7$$

$$= x^2 +10x + 21$$

(ii)We have: $$(4x + 5) (4x + 1)$$

$$=(4x)^2 + (5 + 1)(4x) + 5\times 1$$

$$= 16x^2 +24x + 5$$

(iii)We have: $$(4x - 5) (4x - 1)$$

$$=(4x)^2 - (5 + 1)(4x) + (-5)\times(-1)$$

$$= 16x^2 -24x + 5$$

(iv)We have: $$(4x + 5) (4x - 1)$$

$$=(4x)^2 + (5 - 1)(4x) + (5)\times(-1)$$

$$= 16x^2 +16x - 5$$

(v)We have: $$(2x + 5y) (2x + 3y)$$

$$=(2x)^2 + (5y +3y)(2x) + (5y)\times (3y)$$

$$=(2x)^2 + (8y)(2x) + (5y)\times (3y)$$

$$= 4x^2 +16xy + 15y^2$$

(vi)We have: $$(2a^2 + 9) (2a^2 + 5)$$

$$=(2a^2)^2 + (9+5)(2a^2) + (5)\times (9)$$

$$=(2a^2)^2 + (14)(2a^2) + 45$$

$$= 4a^4 +28a^2 + 45$$

(vii)We have: $$(xyz - 4) (xyz - 2)$$

$$=(xyz)^2 - (4+2)(xyz) + (-4)\times (-2)$$

$$=(xyz)^2 - (6)(xyz) + 8$$

$$= x^2y^2z^2 -6xyz + 8$$