Question

week 7
(x + 3) is a factor of 2x^3 - 5x^2 - 28x + 15. what are the rest of the factors? write your factors in parenthesis. factors: (x + 3)

Explanation:

Step1: Use polynomial long - division

Divide $2x^{3}-5x^{2}-28x + 15$ by $x + 3$.

Step2: Set up long - division

The dividend is $2x^{3}-5x^{2}-28x + 15$ and the divisor is $x + 3$.
First, divide the leading term of the dividend $2x^{3}$ by the leading term of the divisor $x$ to get $2x^{2}$.
Multiply $x + 3$ by $2x^{2}$: $2x^{2}(x + 3)=2x^{3}+6x^{2}$.
Subtract this from the dividend: $(2x^{3}-5x^{2}-28x + 15)-(2x^{3}+6x^{2})=-11x^{2}-28x + 15$.

Step3: Continue long - division

Divide the leading term of the new dividend $-11x^{2}$ by the leading term of the divisor $x$ to get $-11x$.
Multiply $x + 3$ by $-11x$: $-11x(x + 3)=-11x^{2}-33x$.
Subtract this from the new dividend: $(-11x^{2}-28x + 15)-(-11x^{2}-33x)=5x + 15$.

Step4: Final long - division step

Divide the leading term of the new dividend $5x$ by the leading term of the divisor $x$ to get $5$.
Multiply $x + 3$ by $5$: $5(x + 3)=5x+15$.
Subtract this from the new dividend: $(5x + 15)-(5x + 15)=0$.
The quotient is $2x^{2}-11x + 5$.

Step5: Factor the quotient

Factor $2x^{2}-11x + 5$.
We need to find two numbers that multiply to $2\times5 = 10$ and add up to $-11$. The numbers are $-10$ and $-1$.
Rewrite the middle term: $2x^{2}-10x-x + 5$.
Group the terms: $(2x^{2}-10x)-(x - 5)=2x(x - 5)-(x - 5)=(2x - 1)(x - 5)$.

Answer:

$(2x - 1)(x - 5)$