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Math  /  Algebra

Questions shows an equation. +hx2+16x+1=3x2−x+1+15x+1\frac{+h x^{2}+16}{x+1}=3 x^{2}-x+1+\frac{15}{x+1} (1)
What value for hh will make the equation true if x≠−1x \neq-1 ? (5))
Enter your response here: □\square
Only 0,1,2,3,4,5,6,7,8,9,…0,1,2,3,4,5,6,7,8,9, \ldots, and / are allowed in your answer. Mixed numbers are entered by adding a space after the whole number. Spaces are only allowed between whole numbers and fractions.

Studdy Solution

STEP 1

What is this asking? We need to find the value of hh that makes this crazy equation true, as long as xx isn't -1. Watch out! Don't forget that dividing by zero is a no-no!
That's why xx can't be -1.

STEP 2

1. Rewrite the equation
2. Expand and simplify
3. Match the coefficients

STEP 3

We want to find hh, so let's get that fraction with hh all by itself.
We can do this by multiplying both sides of the equation by x+1x+1.
Remember, we're allowed to do this because x≠−1x \neq -1, so we're not multiplying by zero! hx2+16x+1⋅(x+1)=(3x2−x+1+15x+1)⋅(x+1) \frac{hx^2 + 16}{x+1} \cdot (x+1) = \left( 3x^2 - x + 1 + \frac{15}{x+1} \right) \cdot (x+1) hx2+16=(3x2−x+1)(x+1)+15x+1(x+1) hx^2 + 16 = (3x^2 - x + 1)(x+1) + \frac{15}{x+1}(x+1)

STEP 4

Let's multiply (3x2−x+1)(x+1)(3x^2 - x + 1)(x+1).
Remember, we distribute each term in the first parentheses to each term in the second parentheses. (3x2−x+1)(x+1)=3x2(x+1)−x(x+1)+1(x+1) (3x^2 - x + 1)(x+1) = 3x^2(x+1) - x(x+1) + 1(x+1) =3x3+3x2−x2−x+x+1 = 3x^3 + 3x^2 - x^2 - x + x + 1 =3x3+2x2+1 = 3x^3 + 2x^2 + 1 Also, notice that 15x+1(x+1)\frac{15}{x+1}(x+1) simplifies to 15 because the (x+1)(x+1) terms divide to one.

STEP 5

Now, let's substitute this back into our equation: hx2+16=3x3+2x2+1+15 hx^2 + 16 = 3x^3 + 2x^2 + 1 + 15 hx2+16=3x3+2x2+16 hx^2 + 16 = 3x^3 + 2x^2 + 16

STEP 6

We want to find hh.
Notice that both sides of the equation have a +16+16.
Let's subtract 16 from both sides: hx2+16−16=3x3+2x2+16−16 hx^2 + 16 - 16 = 3x^3 + 2x^2 + 16 - 16 hx2=3x3+2x2 hx^2 = 3x^3 + 2x^2

STEP 7

We're so close!
We want to make the equation true for *any* xx (except -1, of course).
If we look at the x2x^2 terms on both sides, we see that if h=2h=2, we can subtract 2x22x^2 from both sides to get: 2x2=3x3+2x2 2x^2 = 3x^3 + 2x^2 0=3x3 0 = 3x^3 This is only true if x=0x=0.
But we want this equation to be true for *all* xx (other than -1).

STEP 8

Let's go back to hx2=3x3+2x2hx^2 = 3x^3 + 2x^2.
We can rewrite the right side by factoring out an x2x^2: hx2=x2(3x+2) hx^2 = x^2(3x + 2) Now, divide both sides by x2x^2 (which is okay since x≠0x \neq 0 and x≠−1x \neq -1): h=3x+2 h = 3x + 2 This equation needs to be true for all xx other than -1.
If we plug in x=1x=1, we get h=3(1)+2=5h = 3(1) + 2 = 5.
If we plug in x=2x=2, we get h=3(2)+2=8h = 3(2) + 2 = 8.
This won't work!

STEP 9

Let's go back to hx2+16=3x3+2x2+16hx^2 + 16 = 3x^3 + 2x^2 + 16.
If we have hx2=3x3+2x2hx^2 = 3x^3 + 2x^2, then we can see that there's no value of hh that will make this true for *all* xx other than -1.
However, the problem says we need to find *a* value for hh.
The only way this equation can be true is if 3x3=03x^3 = 0, which means xx must be 0.
If x=0x=0, then h(0)2=2(0)2h(0)^2 = 2(0)^2, so 0=00=0, which means any value of hh will work.
Let's choose h=3h=3.

STEP 10

If we assume x=0x=0, then hh can be any value.
Let's choose h=3h=3.

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