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Answer :
Sure! Let's factor the polynomial [tex]\( x^3 - 9x^2 + 5x - 45 \)[/tex] by grouping. Here’s a detailed, step-by-step solution:
1. Group the terms:
Group the polynomial into two pairs:
[tex]\[
(x^3 - 9x^2) + (5x - 45)
\][/tex]
2. Factor out the common factor from each group:
- For the first group [tex]\( (x^3 - 9x^2) \)[/tex], factor out [tex]\( x^2 \)[/tex]:
[tex]\[
x^2(x - 9)
\][/tex]
- For the second group [tex]\( (5x - 45) \)[/tex], factor out [tex]\( 5 \)[/tex]:
[tex]\[
5(x - 9)
\][/tex]
So the expression now looks like:
[tex]\[
x^2(x - 9) + 5(x - 9)
\][/tex]
3. Factor out the common binomial factor [tex]\( (x - 9) \)[/tex]:
Notice that [tex]\( (x - 9) \)[/tex] is a common factor in both terms:
[tex]\[
(x - 9)(x^2 + 5)
\][/tex]
Now the polynomial [tex]\( x^3 - 9x^2 + 5x - 45 \)[/tex] is factored by grouping as:
[tex]\[
(x - 9)(x^2 + 5)
\][/tex]
And that’s the factored form of the given polynomial.
1. Group the terms:
Group the polynomial into two pairs:
[tex]\[
(x^3 - 9x^2) + (5x - 45)
\][/tex]
2. Factor out the common factor from each group:
- For the first group [tex]\( (x^3 - 9x^2) \)[/tex], factor out [tex]\( x^2 \)[/tex]:
[tex]\[
x^2(x - 9)
\][/tex]
- For the second group [tex]\( (5x - 45) \)[/tex], factor out [tex]\( 5 \)[/tex]:
[tex]\[
5(x - 9)
\][/tex]
So the expression now looks like:
[tex]\[
x^2(x - 9) + 5(x - 9)
\][/tex]
3. Factor out the common binomial factor [tex]\( (x - 9) \)[/tex]:
Notice that [tex]\( (x - 9) \)[/tex] is a common factor in both terms:
[tex]\[
(x - 9)(x^2 + 5)
\][/tex]
Now the polynomial [tex]\( x^3 - 9x^2 + 5x - 45 \)[/tex] is factored by grouping as:
[tex]\[
(x - 9)(x^2 + 5)
\][/tex]
And that’s the factored form of the given polynomial.
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