multiplication (and multiply 2 + 3 Y) close b y bracket 3 x power 2 minus 4 Y and verify the the resuet for x=1,y=2
Answers
Answer:
SOLUTION
TO DETERMINE
To multiply
\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) }(4x
2
+3y)(3x
2
−4y)
TO verify the result for x = 1 & y = 2
EVALUATION
MULTIPLICATION
\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) }(4x
2
+3y)(3x
2
−4y)
= \sf{4 {x}^{2}(3 {x}^{2} - 4y) + 3y(3 {x}^{2} - 4y)}=4x
2
(3x
2
−4y)+3y(3x
2
−4y)
= \sf{12 {x}^{4} - 16 {x}^{2}y + 9 {x}^{2}y - 12 {y}^{2} }=12x
4
−16x
2
y+9x
2
y−12y
2
= \sf{12 {x}^{4} -7 {x}^{2}y - 12 {y}^{2} }=12x
4
−7x
2
y−12y
2
Hence
\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) = 12 {x}^{4} -7 {x}^{2}y - 12 {y}^{2} }(4x
2
+3y)(3x
2
−4y)=12x
4
−7x
2
y−12y
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} }(4x
2
+3y)(3x
2
−4y)=12x
4
−7x
2
y−12y
2
LHS
\sf{(4 {x}^{2} + 3y)(3 {x}^{2} - 4y) }(4x
2
+3y)(3x
2
−4y)
= \sf{ \bigg[4 \times {(1)}^{2} + 3 \times 2\bigg]\bigg[3 \times {(1)}^{2} - 4 \times 2 \bigg] }=[4×(1)
2
+3×2][3×(1)
2
−4×2]
= (4 + 6)(3 - 8)=(4+6)(3−8)
= 10 \times (- 5)=10×(−5)
= - 50=−50
RHS
\sf{ = 12 {x}^{4} -7 {x}^{2}y - 12 {y}^{2} }=12x
4
−7x
2
y−12y
2
= \sf{ 12 {(1)}^{4} -7 \times {(1)}^{2} \times 2 - 12 \times {(2)}^{2} }=12(1)
4
−7×(1)
2
×2−12×(2)
2
= \sf{ 12 - 14 - 48 }=12−14−48
= - 50=−50
∴ LHS = RHS
Hence verified
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