Question

factor by grouping.
ex
$2x^3 - 10x^2 + 15 - 3x$
$2x^2(x - 5) + 3(5 - x)$ these factors are similar.
$2x^2(x - 5) + 3(-1)(-5 + x)$ factor $(-1)$ from the $2^{nd}$ term.
$2x^2(x - 5) - 3(x - 5)$ commutative property.
$(x - 5)(2x^2 - 3)$ simplify.
1
$4x^3 - 16x^2 - 8 + 2x$
2
$9x^3 - 27x + 21 - 7x^2$

Explanation:

Problem 1: Factor \(4x^{3}-16x^{2}-8 + 2x\) by grouping

Step 1: Group the terms

Group the first two terms and the last two terms: \((4x^{3}-16x^{2})+(-8 + 2x)\)

Step 2: Factor out the GCF from each group

Factor out \(4x^{2}\) from the first group and \(2\) from the second group: \(4x^{2}(x - 4)+2(-4 + x)\)

Step 3: Rewrite the second group to match the first

Notice that \(-4 + x=x - 4\), so we can rewrite the expression as: \(4x^{2}(x - 4)+2(x - 4)\)

Step 4: Factor out the common binomial factor

Factor out \((x - 4)\) from both terms: \((x - 4)(4x^{2}+2)\)

Step 5: Factor out the GCF from the second factor (optional, but to simplify)

Factor out \(2\) from \(4x^{2}+2\): \((x - 4)\times2(2x^{2}+1)=2(x - 4)(2x^{2}+1)\)

Step 1: Combine like terms (group \(x^{3}\) terms)

First, combine the \(x^{3}\) terms: \((9x^{3}-7x^{3})+(-27x + 21)\)
Simplify the \(x^{3}\) terms: \(2x^{3}-27x + 21\) (Wait, maybe a better grouping: Let's group as \((9x^{3}-7x^{3})+(-27x + 21)\) is not helpful. Let's re - group: \((9x^{3}-27x)+(-7x^{3}+21)\)

Step 2: Factor out the GCF from each group

Factor out \(9x\) from the first group and \(-7\) from the second group: \(9x(x^{2}-3)-7(x^{3}-3)\) (Wrong, let's do it again. Correct grouping: \((9x^{3}-7x^{3})+(-27x + 21)\) is not good. Let's group \(9x^{3}-7x^{3}\) and \(-27x + 21\) is not correct. Let's try \((9x^{3}-27x)+(-7x^{3}+21)\)
Factor out \(9x\) from \(9x^{3}-27x\): \(9x(x^{2}-3)\)
Factor out \(-7\) from \(-7x^{3}+21\): \(-7(x^{3}-3)\) (Still wrong). Wait, correct approach:
Let's re - arrange the terms: \(9x^{3}-7x^{3}-27x + 21\)
Combine like terms for \(x^{3}\): \(2x^{3}-27x + 21\) (Not helpful). Let's try another grouping: \((9x^{3}-7x^{3})+(-27x + 21)\) is not good. Let's group \(9x^{3}-27x\) and \(-7x^{3}+21\)
Factor \(9x\) from \(9x^{3}-27x\): \(9x(x^{2}-3)\)
Factor \(-7\) from \(-7x^{3}+21\): \(-7(x^{3}-3)\) (No, that's not matching). Wait, maybe the original expression is \(9x^{3}-27x + 21-7x^{2}\) (maybe a typo). Assuming it's \(9x^{3}-7x^{2}-27x + 21\) (more likely). Let's solve this corrected version.

Step 1: Group the terms

Group as \((9x^{3}-7x^{2})+(-27x + 21)\)

Step 2: Factor out the GCF from each group

Factor out \(x^{2}\) from the first group and \(-3\) from the second group: \(x^{2}(9x - 7)-3(9x - 7)\)

Step 3: Factor out the common binomial factor

Factor out \((9x - 7)\): \((9x - 7)(x^{2}-3)\)

If we assume the original problem is \(9x^{3}-27x + 21-7x^{3}\) (with a typo in the exponent of the last \(x\) term, should be \(x^{2}\) instead of \(x^{3}\)), the answer is \((9x - 7)(x^{2}-3)\)

If we take the problem as given \(9x^{3}-27x + 21-7x^{3}\), we can rewrite it as \(2x^{3}-27x + 21\), which is a cubic equation. But factoring by grouping for the given expression (as written) is not straightforward. However, if we assume a typo and the last term is \(-7x^{2}\) instead of \(-7x^{3}\), we get \((9x - 7)(x^{2}-3)\)

Since the original problem might have a typo, but based on the pattern of factoring by grouping, the most probable intended problem (correcting the last \(x^{3}\) to \(x^{2}\)) gives the answer \((9x - 7)(x^{2}-3)\)

Answer:

\(2(x - 4)(2x^{2}+1)\)

Problem 2: Factor \(9x^{3}-27x + 21-7x^{3}\) by grouping