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How do you factor the expression 16b2+60b100 ?

Answer
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Hint: The given equation 16b2+60b100 is a quadratic equation. To factorize a quadratic equation ax2+bx+c we have to write bx as mx + nx such that product of m and n is equal to product of ac. In the equation 16b2+60b100 we have to write 60b = mb + nb such that mn is equal to – 1600. We can evaluate m is 80 and n is – 20.

Complete step by step solution:
The given equation is 16b2+60b100
If we compare it with ax2+bx+c we get a = 16, b = 60 and c = - 100
The value of ac is equal to – 1600
So, we have to find 2 numbers such that their sum is 60 and product is – 1600. The numbers are 80 and – 20
So, we can write
 16b2+60b100=16b2+80b20b100
We can take 16b common from first 2 terms and – 20 common from last 2 terms
16b2+60b100=16b(b+5)20(b+5)
Now we can take b + 5 common from the whole equation
16b2+60b100=(16b20)(b+5)
Now we can take 4 common from 16b – 20
16b2+60b100=4(4b5)(b+5)

4(4b5)(b+5) is the factored form of the equation 16b2+60b100.

Note: The graph of all quadratic equations is a parabola whose axis is always parallel to y axis. In quadratic equation ax2+bx+c , if a is greater than 0 then it will be upward parabola and if a is less than 0 then graph will be downward parabola. If the discriminant (b24ac ) is less than 0 then the parabola will never touch the x axis.
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