Question

multiply (4x^2+3y)by(3x^2-4y) and verify the result for x=1,y=2

Answer 1

SOLUTION

TO DETERMINE

• To multiply

$\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) }$

• TO verify the result for x = 1 & y = 2

EVALUATION

MULTIPLICATION

$\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) }$

$= \sf{4 {x}^{2}(3 {x}^{2} - 4y) + 3y(3 {x}^{2} - 4y)}$

$= \sf{12 {x}^{4} - 16 {x}^{2}y + 9 {x}^{2}y - 12 {y}^{2} }$

$= \sf{12 {x}^{4} -7 {x}^{2}y - 12 {y}^{2} }$

Hence

$\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) = 12 {x}^{4} -7 {x}^{2}y - 12 {y}^{2} }$

VERIFICATION

We need to verify for x = 1 & y = 2 the below result

$\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) = 12 {x}^{4} -7 {x}^{2}y - 12 {y}^{2} }$

LHS

$\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) }$

$= \sf{ \bigg[4 \times {(1)}^{2} + 3 \times 2\bigg]\bigg[3 \times {(1)}^{2} - 4 \times 2 \bigg] }$

$= (4 + 6)(3 - 8)$

$= 10 \times (- 5)$

$= - 50$

RHS

$\sf{ = 12 {x}^{4} -7 {x}^{2}y - 12 {y}^{2} }$

$= \sf{ 12 {(1)}^{4} -7 \times {(1)}^{2} \times 2 - 12 \times {(2)}^{2} }$

$= \sf{ 12 - 14 - 48 }$

$= - 50$

∴ LHS = RHS

Hence verified

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Answer 2

Given :

The data is given as, $\[\left( {4{x^2} + 3y} \right) \times \left( {3{x^2} - 4y} \right)\]$.

To Find:

We have to find the result of $\[\left( {4{x^2} + 3y} \right) \times \left( {3{x^2} - 4y} \right)\]$ and then verify the answer for x=1, y=2.

Solution:

• First, we will multiply the two polynomials,

        $\[\left( {4{x^2} + 3y} \right) \times \left( {3{x^2} - 4y} \right)\]$            .....L.H.S

         $\[\begin{array}{l} \Rightarrow 12{x^4} - 16{x^2}y + 9{x^2}y - 12{y^2}\\\\ \Rightarrow 12{x^4} - 7{x^2}y - 12{y^2}\end{array}\]$....R.H.S

• Secondly, we need to verify the result for x=1 and y=2

        L.H.S

       Substitute the values of  x=1 and y=2 in LHS:

       Then we get,

                       $\[\begin{array}{l} \Rightarrow \left( {4{x^2} + 3y} \right) \times \left( {3{x^2} - 4y} \right)\\\\ \Rightarrow \left( {4 \times {1^2} + 3 \times 2} \right) \times \left( {3 \times {1^2} - 4 \times 2} \right)\\\\ \Rightarrow 10 \times - 5\\\\ \Rightarrow - 50\end{array}\]$

        R.H.S

       Substitute the values of x=1 and y=2 in RHS:

       Then we get,  

                       $\[\begin{array}{l} \Rightarrow 12{x^4} - 16{x^2}y + 9{x^2}y - 12{y^2}\\\\ \Rightarrow 12{x^4} - 7{x^2}y - 12{y^2}\\\\ \Rightarrow 12 \times {1^4} - 7 \times {1^2} \times 2 - 12 \times {2^2}\\\\ \Rightarrow - 50\end{array}\]$

   ⇒ LHS = RHS. Hence, the answer is verified.