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2 years ago

Multiply polynomial (x+6) (x-4)

2 answers:

8 0

Use foil or just plain distribute

if you want foil

FOIL
First
Outer
Inner
Last
that is order to multiply dem numbers

(x+6)(x-4)
First: x times x=x²
Outer: x times -4=-4x
Inner: 6 times x=6x
Last: 6 times -4=-24

add them all together

x²-4x+6x-24
x²+2x-24

Paul [167]2 years ago

3 0

To do this, you use the FOIL method.

F: front
x*x=x^2

O: outer
x*-4=-4x

I: inner
6*x=6x

L: last
6*-4=-24

Add all those up and you get:
x^2-4x+6x-24

Which is equal to:
x^2+2x-24

Hope this helped!

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Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

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Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

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$140x - x^2 - 2xy = 0$

---------------------------------------------------

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

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$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

---------------------------------------------------

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

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