Question

practice problems name
1 what are the points of... g(x)=x+1? -1,0 (1,0)

solve, w/ ÷ *

to also check.
this workd
5 from unit 2, lesson 5
what is the value of 6(x - 2)(x - 3) + 4(x - 2)(x - 5) when x = -3?
6(-3 - 2)(-3 - 3) + 4(-3 - 2)(-3 - 5
6 from unit 2, lesson 6
match each polynomial function with its leading coefficient.
a ( p(x) = (x + 2)(2x - 3)(4x + 7) )
1 40
b ( p(x) = \frac{1}{2}(x - 2)(2x - 3)(4x + 7) )
2 8
c ( p(x) = 5(x - 2)(2x - 3)(4x + 7) )
3 4
d ( p(x) = -(x - 2)(2x - 3)(4x + 7) )
4 2
e ( p(x) = \frac{1}{4}(x + 2)(2x - 3)(4x + 7) )
5 -8
math prs
10x from

Explanation:

Step1: Factor out common terms

Notice that both terms have a common factor of \((x - 2)\). So we can factor the expression:

$$
LATEXBLOCK0
$$

Step2: Simplify the bracket

First, expand the terms inside the bracket:

$$
LATEXBLOCK1
$$

So the expression becomes \((x - 2)(10x - 38)\)

Step3: Substitute \(x=-3\)

Now substitute \(x = - 3\) into the factored expression:
First, calculate \(x-2\) when \(x=-3\): \(-3 - 2=-5\)
Then, calculate \(10x-38\) when \(x = - 3\): \(10\times(-3)-38=-30 - 38=-68\)
Now multiply these two results: \((-5)\times(-68) = 340\)

Alternatively, we can substitute \(x=-3\) directly into the original expression:

Step1: Substitute \(x = - 3\) into each term

For the first term \(6(x - 2)(x - 3)\):
\(x=-3\), so \(x - 2=-3-2=-5\), \(x - 3=-3 - 3=-6\)
Then \(6\times(-5)\times(-6)=6\times30 = 180\)
For the second term \(4(x - 2)(x - 5)\):
\(x=-3\), so \(x - 2=-5\), \(x - 5=-3-5=-8\)
Then \(4\times(-5)\times(-8)=4\times40 = 160\)

Step2: Add the two terms

Now add the two results: \(180+160 = 340\)

Answer:

\(340\)