What are the possible rational zeros of f(x) = x4 + 6x3 – 3x2 + 17x – 15?.

1 answer:

4 0

$f(x) = x^4 + 6x^3-3x^2 + 17x-15\\\\15:\{\pm1;\ \pm3;\ \pm5;\ \pm15\}\\1:\{\pm1\}\\\\Answer:\boxed{\{\pm1;\ \pm3;\ \pm5;\ \pm15\}}$

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Which will result in a difference of squares? (–7x + 4)(–7x + 4) (–7x + 4)(4 – 7x) (–7x + 4)(–7x – 4) (–7x + 4)(7x – 4)

maw [93]

You can simply expand each product and see whether it gives you a difference of squares.

• $\mathsf{(-7x+4)\cdot (-7x+4)}$

That's actually $\mathsf{(-7x+4)^2:}$

$\mathsf{(-7x+4)^2}\\\\ \mathsf{=(-7x+4)\cdot (-7x+4)}\\\\ \mathsf{=(-7x+4)\cdot (-7x)+(-7x+4)\cdot 4}\\\\ \mathsf{=49x^2-28x-28x+16}$

$\mathsf{=49x^2-56x+16}$ ✖

which is not a difference of squares.

—————

• $\mathsf{(-7x+4)\cdot (4-7x)}$

$\mathsf{=(-7x+4)\cdot 4-(-7x+4)\cdot 7x}\\\\ \mathsf{=-28x+16-(-49x^2+28x)}\\\\ \mathsf{=-28x+16+49x^2-28x}$

$\mathsf{=49x^2-56x+16}$ ✖

which is not a difference of squares.

—————

• $\mathsf{(-7x+4)\cdot (-7x-4)}$

$\mathsf{=(-7x+4)\cdot (-7x)-(-7x+4)\cdot 4}\\\\ \mathsf{=49x^2-28x-(-28x+16)}\\\\ \mathsf{=49x^2-\diagup\!\!\!\!\! 28x+\diagup\!\!\!\!\! 28x-16}\\\\ \mathsf{=49x^2-16}$

$\mathsf{=(7x)^2-4^2}$ ✔

That is a difference of two squares.

—————

• $\mathsf{(-7x+4)\cdot (7x-4)}$

$\mathsf{=(-7x+4)\cdot 7x-(-7x+4)\cdot 4)}\\\\ \mathsf{=-49x^2+28x-(-28x+16)}\\\\ \mathsf{=-49x^2+28x+28x-16}$

$\mathsf{=-49x^2+56x-16}$ ✖

which is not a difference of squares.

—————

Only the third option will result in a difference of squares.

Answer: (− 7x + 4) · (− 7x − 4).

I hope this helps. =)

8 0

3 years ago

Read 2 more answers

Square Roots

dlinn [17]

4 is the square root of 16

i think that what you are asking, but if it is not tell me and i can try to help more :)

4 0

3 years ago

John has seven grapes he give 5 to jack and 1 to hailey how many grapes does he have left

weeeeeb [17]

He as 1 grape left
Since he had 7 and gave 6 away :)

3 0