Q. \(\left(x^2+2x+5\right)-\left(-2x^2+6\right)\)

Q: Simplify (x^2+2x+5)-(-2x^2+6) .

(x^2+2x+5)-(-2x^2+6)=x^2+2x+5+2x^2-6=3x^2+2x-1 .

Q: What happens when subtracting a negative polynomial -(-2x^2+6) ?

Distribute the minus sign: -(-2x^2+6)=+2x^2-6 .

Q: How do you combine like terms after expansion?

Combine x^2-terms: x^2+2x^2=3x^2, keep 2x, and constants: 5-6=-1. Final: 3x^2+2x-1.

Q: What is the value at x=1 for the simplified expression?

Substitute into 3x^2+2x-1: 3(1)^2+2(1)-1=3+2-1=4.

Q: What is the value at x=0 for (x^2+2x+5)-(-2x^2+6) ?

Substitute directly: (0+0+5)-(-0+6)=5-6=-1.

Q: Can you expand first without simplifying signs incorrectly?

Yes: x^2+2x+5-(-2x^2+6)=x^2+2x+5+2x^2-6, then combine to get 3x^2+2x-1.

Answer

Simplify the expression by distributing the minus sign:

\[
(x^2+2x+5)-(-2x^2+6)=(x^2+2x+5)+2x^2-6
\]

Combine like terms:

\[
( x^2+2x^2 ) + 2x + (5-6)=3x^2+2x-1
\]

Final result: \(3x^2+2x-1\)

Detailed Explanation

We want to simplify the expression:

\[
(x^2+2x+5)-(-2x^2+6).
\]

Step 1: Distribute the subtraction sign.

The expression has a minus sign in front of the parentheses. That means every term inside the parentheses changes sign.

So:

\[
-( -2x^2 + 6 ) = +2x^2 – 6.
\]

Now rewrite the whole expression by replacing the subtracted parentheses with its sign-changed form:

\[
(x^2+2x+5) + 2x^2 – 6.
\]

Step 2: Combine like terms.

Now group terms with the same variable powers:

Quadratic terms (in \(x^2\)): \(x^2 + 2x^2\)
Linear terms (in \(x\)): \(2x\)
Constant terms: \(5 – 6\)

Combine each group:

\[
x^2 + 2x^2 = 3x^2,
\]

\[
5 – 6 = -1.
\]

So the expression becomes:

\[
3x^2 + 2x – 1.
\]

Final Answer:

\[
3x^2 + 2x – 1.
\]

See full solution

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Algebra FAQ

Simplify \( (x^2+2x+5)-(-2x^2+6) \).

\( (x^2+2x+5)-(-2x^2+6)=x^2+2x+5+2x^2-6=3x^2+2x-1 \).

What happens when subtracting a negative polynomial \( -(-2x^2+6) \)?

Distribute the minus sign: \( -(-2x^2+6)=+2x^2-6 \).

How do you combine like terms after expansion?

Combine \(x^2\)-terms: \(x^2+2x^2=3x^2\), keep \(2x\), and constants: \(5-6=-1\). Final: \(3x^2+2x-1\).

What is the value at \(x=1\) for the simplified expression?

Substitute into \(3x^2+2x-1\): \(3(1)^2+2(1)-1=3+2-1=4\).

What is the value at \(x=0\) for \( (x^2+2x+5)-(-2x^2+6) \)?

Substitute directly: \((0+0+5)-(-0+6)=5-6=-1\).

Can you expand first without simplifying signs incorrectly?

Yes: \(x^2+2x+5-(-2x^2+6)=x^2+2x+5+2x^2-6\), then combine to get \(3x^2+2x-1\).

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