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Multiplying Two Binomials by a Trinomial
- Thread starter Jason76
- Start date
D
Deleted member 4993
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Is this right?
\(\displaystyle (a + b)(c + d)(e + f + g)\)
\(\displaystyle (ae + af + ag)(be + bf + bg)\)
Is this headed in the right direction?
\(\displaystyle (a + b)(c + d)(e + f + g)\)
\(\displaystyle = [a* (e + f + g) + b*(e + f + g)](c + d)\)
\(\displaystyle = [ae + af + ag + be + bf + bg](c + d)\)
and continue.....
HallsofIvy
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- Jan 27, 2012
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Is what right? You have two expressions on two lines with no indication as to what is supposed to be true of them. If you mean "are they equal?", the most common assumption in such a situation, the answer is no. What is true is that \(\displaystyle (a+ b)(e+ f+ g)= ae+ naf+ ag+ be+ bf+ bg\). That is, the two are added, not multiplied, and "(c+ d)" has nothing to do with it.Is this right?
\(\displaystyle (a + b)(c + d)(e + f + g)\)
\(\displaystyle (ae + af + ag)(be + bf + bg)\)
Is this headed in the right direction?
Is what right? You have two expressions on two lines with no indication as to what is supposed to be true of them. If you mean "are they equal?", the most common assumption in such a situation, the answer is no. What is true is that \(\displaystyle (a+ b)(e+ f+ g)= ae+ naf+ ag+ be+ bf+ bg\). That is, the two are added, not multiplied, and "(c+ d)" has nothing to do with it.
So (c + d) has nothing to do with it. Could you FOIL the the binomials and then multiply them to the trinomial? Multiplying two trinomials or one binomial by a trinomial is understood.
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So (c + d) has nothing to do with it. Could you FOIL the the binomials and then multiply them to the trinomial? Multiplying two trinomials or one binomial by a trinomial is understood.
Yup.