Factor completely x^2+9x+18 and Expand (a+3)^2
1 answer:
• Factor x² + 9x + 18 completely:
In order to make the process easier, I'm going to try to find two integer numbers, so that
their sum is 9
their product is 18
Just search into the set of the divisors of 18
D(18) = {1, 2, 3, 6, 9, 18}
Take the numbers 3 and 6, and you verify that
3 + 6 = 9 <span>✔
</span>
3 × 6 = 18 <span>✔
</span>
So, rewrite conveniently 9x as 3x + 6x, and then factor the quadratic polynomial by grouping:
x² + 9x + 18
= x² + 3x + 6x + 18
= x² + 3x + 6x + 6 · 3
= x · (x + 3) + 6 · (x + 3)
= (x + 3) · (x + 6) ✔
and that is the complete factorization of x² + 9x + 18.
_______________
• Expand (a + 3)²:
Just apply the distributive property of multiplication, and you have
(a + 3)²
= (a + 3) · (a + 3)
= (a + 3) · a + (a + 3) · 3
= a · a + 3 · a + a · 3 + 3 · 3
= a² + 3a + 3a + 3²
= a² + 6a + 9 <——<span>—</span> this is the expanded form.
I hope this helps. =)
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