To express the product of \(2x^2 + 6x - 8\) and \(x + 3\) in standard form, we will use the distributive property (also known as the FOIL method for binomials).
Start by distributing each term in the polynomial \(2x^2 + 6x - 8\) by each term in the binomial \(x + 3\):
-
Multiply \(2x^2\) by \(x\): \[ 2x^2 \cdot x = 2x^3 \]
-
Multiply \(2x^2\) by \(3\): \[ 2x^2 \cdot 3 = 6x^2 \]
-
Multiply \(6x\) by \(x\): \[ 6x \cdot x = 6x^2 \]
-
Multiply \(6x\) by \(3\): \[ 6x \cdot 3 = 18x \]
-
Multiply \(-8\) by \(x\): \[ -8 \cdot x = -8x \]
-
Multiply \(-8\) by \(3\): \[ -8 \cdot 3 = -24 \]
Now, combine all these products: \[ 2x^3 + 6x^2 + 6x^2 + 18x - 8x - 24 \]
Next, combine like terms:
- The \(x^2\) terms: \(6x^2 + 6x^2 = 12x^2\)
- The \(x\) terms: \(18x - 8x = 10x\)
Putting it all together, we have: \[ 2x^3 + 12x^2 + 10x - 24 \]
Thus, the product of \(2x^2 + 6x - 8\) and \(x + 3\) in standard form is: \[ \boxed{2x^3 + 12x^2 + 10x - 24} \]
2 answers