Find the factors of f(x) given that x = 5 is a zero.
1 answer:
- If x = 5 is a zero of the given polynomial, then,
- (x - 5) is a factor of the polynomial. [Since, x = 5 or, x - 5 = 0]
- Now, divide the polynomial with (x - 5) using long division method. (See the picture)
- We get (x^2 - x - 6) as the quotient.
- Now, factorise the above polynomial:
- (x^2 - x - 6)
- = (x^2 - 3x + 2x - 6)
- = x(x - 3) + 2(x - 3)
- = (x - 3)(x + 2)
- Therefore, x^3 − 6x^2 − x + 30 = (x − 5)(x − 3)(x + 2)
<u>Answer</u><u>:</u>
<u>B.</u><u>(x − 5)(x − 3)(x + 2)</u>
Hope you could get an idea from here.
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Check the picture below.
notice the sides, now, on the second triangle, side 6 slants a bit more to fit in 13, on the third triangle, side 6 slants even further to fit 13 in, now, if 6 were to slant completely, it'll make a flat-line with side 5, and there will be a triangle no more.
but even if side 6 would stretch to a flat-line, 5+6 is just 11, whilst side 13 is longer than that, so no dice.
notice the sides, now, on the second triangle, side 6 slants a bit more to fit in 13, on the third triangle, side 6 slants even further to fit 13 in, now, if 6 were to slant completely, it'll make a flat-line with side 5, and there will be a triangle no more.
but even if side 6 would stretch to a flat-line, 5+6 is just 11, whilst side 13 is longer than that, so no dice.
Answer:
The rocket will reach its maximum height after 6.13 seconds
Step-by-step explanation:
To find the time of the maximum height of the rocket differentiate the equation of the height with respect to the time and then equate the differentiation by 0 to find the time of the maximum height
∵ y is the height of the rocket after launch, x seconds
∵ y = -16x² + 196x + 126
- Differentiate y with respect to x
∴ y' = -16(2)x + 196
∴ y' = -32x + 196
- Equate y' by 0
∴ 0 = -32x + 196
- Add 32x to both sides
∴ 32x = 196
- Divide both sides by 32
∴ x = 6.125 seconds
- Round it to the nearest hundredth
∴ x = 6.13 seconds
∴ The rocket will reach its maximum height after 6.13 seconds
There is another solution you can find the vertex point (h , k) of the graph of the quadratic equation y = ax² + bx + c, where h = and k is the value of y at x = h and k is the maximum/minimum value
∵ a = -16 , b = 196
∴
∴ h = 6.125
∵ h is the value of x at the maximum height
∴ x = 6.125 seconds
- Round it to the nearest hundredth
∴ x = 6.13 seconds