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

Answer

Multiply and combine like terms:
\[
(x-1)(x^2+3x-5)=x(x^2+3x-5)-1(x^2+3x-5)
\]
\[
= x^3+3x^2-5x -x^2-3x+5 = x^3+2x^2-8x+5.
\]

Final result: \(\;x^3+2x^2-8x+5.\)

Detailed Explanation

Problem: Expand the product \( (x-1)(x^2+3x-5) \).

Step 1 — Apply the distributive property (multiply each term of the second polynomial by each term of the first):

Write the product as a sum of two products:
\( (x-1)(x^2+3x-5) = x\,(x^2+3x-5) – 1\,(x^2+3x-5) \).
Step 2 — Multiply the first term \(x\) by each term of the second polynomial:

Compute:
\( x\cdot x^2 = x^3 \),
\( x\cdot 3x = 3x^2 \),
\( x\cdot (-5) = -5x \).

So \( x\,(x^2+3x-5) = x^3 + 3x^2 – 5x \).
Step 3 — Multiply the second term \(-1\) by each term of the second polynomial:

Compute:
\( -1\cdot x^2 = -x^2 \),
\( -1\cdot 3x = -3x \),
\( -1\cdot (-5) = +5 \).

So \( -1\,(x^2+3x-5) = -x^2 – 3x + 5 \).
Step 4 — Add the two results and combine like terms:

Sum:
\( (x^3 + 3x^2 – 5x) + (-x^2 – 3x + 5) \).

Group like terms:
Cubic term: \( x^3 \).
Quadratic terms: \( 3x^2 + (-x^2) = 2x^2 \).
Linear terms: \( -5x + (-3x) = -8x \).
Constant term: \( 5 \).

Thus the combined result is
\( x^3 + 2x^2 – 8x + 5 \).
Final Answer:

\( x^3 + 2x^2 – 8x + 5 \)

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FAQs

How do I expand \((x-1)(x^2+3x-5)\)?

-A: Distribute: \(x(x^2+3x-5)-1(x^2+3x-5)\). Combine terms to get \(x^3+2x^2-8x+5\).

What is the fully expanded form?

-A: \((x-1)(x^2+3x-5)=x^3+2x^2-8x+5\).

What is the degree and leading coefficient?

-A: Degree is \(3\). Leading coefficient is \(1\) (from \(x^3\)).

Can it be factored further over the reals?

-A: Yes: original factorization is \((x-1)(x^2+3x-5)\). The quadratic has roots \(\frac{-3\pm\sqrt{29}}{2}\), so fully over \(\mathbb{R}\): \((x-1)\left(x-\frac{-3+\sqrt{29}}{2}\right)\left(x-\frac{-3-\sqrt{29}}{2}\right)\).

Solve \((x-1)(x^2+3x-5)=0\).

-A: Solutions are \(x=1\) and \(x=\dfrac{-3\pm\sqrt{29}}{2}\).

What is the y-intercept?

-A: Set \(x=0\): value is \((0-1)(-5)=5\). So y-intercept is \((0,5)\).

What are the x-intercepts?

-A: The x-intercepts are the real roots: \(x=1\) and \(x=\dfrac{-3\pm\sqrt{29}}{2}\).

What is the value at \(x=2\)?

-A: \((2-1)(2^2+3\cdot2-5)=1\cdot5=5\). So \(f(2)=5\).

What is the end behavior (graph behavior as \(x\to\pm\infty\))?

-A: Leading term \(x^3\): as \(x\to\infty\), \(f(x)\to\infty\); as \(x\to-\infty\), \(f(x)\to-\infty\).

What is the derivative?

-A: Differentiate \(x^3+2x^2-8x+5\): \(f'(x)=3x^2+4x-8\).

Where are the critical points (local extrema)?

-A: Solve \(3x^2+4x-8=0\): \(x=\dfrac{-4\pm\sqrt{112}}{6}=\dfrac{-2\pm2\sqrt{7}}{3}\) (two real critical points).

Can the quadratic \(x^2+3x-5\) be factored over integers/rationals?

-A: No. Discriminant \( \Delta=9+20=29\) is not a perfect square, so it has irrational roots and no rational integer factorization.

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