Use suitable identities to find the following products:
(i) $ (x+4)(x+10) $
(ii) $ (x+8)(x-10) $
(iii) $ (3 x+4)(3 x-5) $
(iv) $ \left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) $
(v) $ (3-2 x)(3+2 x) $

AcademicMathematicsNCERTClass 9

To do:

We have to use suitable identities to find the given products.

Solution:

We know that,

$(p + a)(p+b) = p( p+ b) + a(p+b)$

$=p^2+pb+ap+ab$

$=p^2+p(a+b)+ab$

Therefore,

(i) \( (x+4)(x+10) \)

Here, $p=x, a=4$ and $b=10$

This implies,

$(x+4)(x+10) = x^2+x(4+10)+4\times10$

$=x^2+14x+40$

Hence, $(x+4)(x+10)=x^2+14x+40$.

(ii) \( (x+8)(x-10) \)

Here, $p=x, a=8$ and $b=-10$

This implies,

$(x+8)(x-10) = x^2+x(8-10)+8\times(-10)$

$=x^2-2x-80$

Hence, $(x+8)(x-10)=x^2-2x-80$.

(iii) \( (3 x+4)(3 x-5) \)

Here, $p=3x, a=4$ and $b=-5$

This implies,

$(3x+4)(3x-5) = (3x)^2+3x(4-5)+4\times(-5)$

$=9x^2+3x(-1)-20$

$=9x^2-3x-20$

Hence, $(3x+4)(3x-5)=9x^2-3x-20$.

(iv) \( \left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right) \)

Here, $p=y^2, a=\frac{3}{2}$ and $b=-\frac{3}{2}$

This implies,

$(y^2+\frac{3}{2})(y^2-\frac{3}{2}) = (y^2)^2+y^2(\frac{3}{2}-\frac{3}{2})+\frac{3}{2}\times(-\frac{3}{2})$

$=y^4+y^2(0)-(\frac{3}{2})^2$

$=y^4-0-\frac{9}{4}$

$=y^4-\frac{9}{4}$

Hence, $(y^2+\frac{3}{2})(y^2-\frac{3}{2})=y^4-\frac{9}{4}$.

(v) \( (3-2 x)(3+2 x) \)

Here, $p=3, a=-2x$ and $b=2x$

This implies,

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

$=9+3(0)-4x^2$

$=-4x^2+9$

Hence, $(3-2x)(3+2x)=-4x^2+9$.

raja
Updated on 10-Oct-2022 13:39:07

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