Question #80d85

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Question #80d85

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Answer 1 · 2017-02-19T01:07:55

$p (a - 1) = 2 \cdot (1 - a)$ $p(a-1) = 2 * (1-a)$

Explanation:

It's given
$p (a) = - 3 a + a$ $p(a) = -3a + a$

Let's simplify the given polynomial in one variable a little more so as to make our calculations easy.

$p (a) = - 3 a + a$ $p(a) = -3a + a$
$p (a) = a \cdot (- 3 + 1)$ $p(a) = a * (-3 + 1)$
$p (a) = a \cdot (- 2)$ $p(a) = a * (-2)$
$p (a) = - 2 a$ $p(a) = -2a$

Now, they have asked us to find $p (a - 1)$ $p(a-1)$ . To do this; just plug in $(a - 1)$ $(a-1)$ in place of $a$ $a$ in the given polynomial.

Therefore,

$p (a - 1) = (- 2) \cdot (a - 1)$ $p(a-1) = (-2) * (a-1)$
$p (a - 1) = (- 2) (a) - (- 2) (1)$ $p(a-1) = (-2)(a) - (-2)(1)$
$p (a - 1) = - 2 a + 2$ $p(a-1) = -2a + 2$

Rearranging the equation,

$p (a - 1) = 2 - 2 a$ $p(a-1) = 2 -2a$
$p (a - 1) = 2 \cdot (1 - a)$ $p(a-1) = 2 * (1-a)$

And there you have your final answer!

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Question #80d85

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