Question

Use the suitable identity and find their products.



(i) (4x+5)(4x-1)

(ii) (2a2+9)(2a2+5)​

Answer 1

Solution

Q.1 Expand.

We could use this identity: $(x+a)(x+b)=x^2+(a+b)x+ab$

Here x was replaced by 4x, a by 5, and b by -1.

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

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

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

Q.2 Expand.

Using the same identity, this time we replace them with different values.

Here x was replaced by 2a², a by 9, b by 5.

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

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

More information

Identity comes in handy when we expand. Also, if we do the process backward it is called factorization.

For example, let's use one of our solutions.

What will the factorization of $4a^4+28a^2+45$ be? It is $(2a^2+9)(2a^2+5)$.

I hope you understood!

Answer 2

1]

$\tt (4x + 5)(4x-1)$

Apply identity

$\tt(x + a)(x+b)= x^{2} +(a+b)x+ab$

Putting the value together

$\sf\bigg( 4x\bigg)^{2} +(5-1)4x +5 \times(-1)$

$\sf \bigg(4x^{2} \bigg)+(4)4x + -5$

$\sf 16x^{2} +16x-5$

2]

Using identity

$\tt(x + a)(x+b)= x^{2} +(a+b)x+ab$

$\sf \bigg(2a\bigg)^{2} + (9+5)\times2a^{2} + 9 \times 5$

$\sf 4a^{4} + (14) \times 2a^{2} + 45$

$\sf 4a^{4} + 28a^{2}+45$