Q.
\(x^2+x-2\)

Q: How do you factor x^2+x-2 using the ac method?

a=1,c=-2. Find numbers multiplying to -2 and adding to 1: 2 and -1. So x^2+ x-2=(x+2)(x-1).

Q: What is the product ac and the target sum for factoring x^2+x-2?

ac=1\cdot (-2)=-2. Target sum is the middle coefficient 1.

Q: Solve x^2+x-2=0 by setting factors to zero.

From (x+2)(x-1)=0, get x+2=0 so x=-2, and x-1=0 so x=1.

Q: How can you check the factorization (x+2)(x-1) quickly?

Expand: (x+2)(x-1)=x^2-x+2x-2=x^2+x-2. Matches the original.

Q: Can x^2+x-2 be factored using integer roots (rational root theorem)?

Test possible roots \pm1,\pm2. 1 and -2 make the polynomial zero, giving factors (x-1)(x+2).

Answer

We factor \(x^2+x-2\) by finding two numbers that multiply to \(-2\) and add to \(1\). Those numbers are \(2\) and \(-1\).

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

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

Detailed Explanation

We want to factor the quadratic polynomial

\[
x^2 + x – 2.
\]

Step 1: Find two numbers.

We need two integers \(a\) and \(b\) such that

\(a \cdot b = -2\) (so their product matches the constant term \(-2\)), and
\(a + b = 1\) (so their sum matches the coefficient of \(x\), which is \(1\)).

Step 2: List number pairs for \(-2\).

The integer pairs whose product is \(-2\) are:

\(2\) and \(-1\), because \(2 \cdot (-1) = -2\).
\(-2\) and \(1\), because \((-2) \cdot 1 = -2\).

Step 3: Check which pair sums to \(1\).

\(2 + (-1) = 1\) works.
\((-2) + 1 = -1\) does not.

So the numbers are \(2\) and \(-1\).

Step 4: Write the factorization.

For a quadratic \(x^2 + x – 2\), using \(2\) and \(-1\) gives:

\[
x^2 + x – 2 = (x + 2)(x – 1).
\]

Answer:

\[
(x + 2)(x – 1).
\]

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Algebra FAQ

What are the factors of \(x^2+x-2\) ?

\((x+2)(x-1)\)

How do you factor \(x^2+x-2\) using the \(ac\) method?

\(a=1,c=-2\). Find numbers multiplying to \(-2\) and adding to \(1\): \(2\) and \(-1\). So \(x^2+ x-2=(x+2)(x-1)\).

What is the product \(ac\) and the target sum for factoring \(x^2+x-2\)?

\(ac=1\cdot (-2)=-2\). Target sum is the middle coefficient \(1\).

Solve \(x^2+x-2=0\) by setting factors to zero.

From \((x+2)(x-1)=0\), get \(x+2=0\) so \(x=-2\), and \(x-1=0\) so \(x=1\).

How can you check the factorization \( (x+2)(x-1)\) quickly?

Expand: \((x+2)(x-1)=x^2-x+2x-2=x^2+x-2\). Matches the original.

Can \(x^2+x-2\) be factored using integer roots (rational root theorem)?

Test possible roots \(\pm1,\pm2\). \(1\) and \(-2\) make the polynomial zero, giving factors \((x-1)(x+2)\).

Solve the equation step by step.
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Q. \(x^2+x-2\)