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Q.

[[1]] is polynomial P as a product of linear factors: P(x) = $x^{4} + 4 x^{3} + 6 x^{2} + 4 x - 15$

[Hint 1 and−3 are zeroes of P]

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Detailed Solution

We have been given a polynomial P(x) =

x^{4} + 4 x^{3} + 6 x^{2} + 4 x - 15

.
We have to find the linear factors of the given polynomial.
We have given a hint in the question that 1 and -3 are zeros of a given polynomial so (x−1) & (x+3) are two factors of the polynomial.
Now, to find the other factors we will divide the given polynomial
P(x) =

x^{4} + 4 x^{3} + 6 x^{2} + 4 x - 15

by (x−1) (x+3)
Let us divide the polynomial by using long division method, we have
⇒(x−1)(x+3)=

x^{2} + 2 x - 3

x^{2} + 2 x + 5

\underset{̲}{x^{2} + 2 x - 3 ⟌ x^{4} + 4 x^{3} + 6 x^{2} + 4 x - 15 x^{4} + 2 x^{3} - 3 x^{2}___________________2 x^{3} + 9 x^{2} + 4 x 2 x^{3} + 4 x^{2} - 6 x__________________{5 x}^{2} + 10 x - 15 {5 x}^{2} + 10 x - 15 0}

On dividing, we get quotient

x^{2} + 2 x + 5

and remainder zero.
Now we factorize the obtained quotient by using quadratic formula

x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

.
We get
⇒

x = \frac{- 2 \pm \sqrt{2^{2} - 4 \times 1 \times 5}}{2}

.
⇒

x = \frac{- 2 \pm \sqrt{4 - 20}}{2}

.
⇒

x = \frac{- 2 \pm \sqrt{- 16}}{2}

.
⇒

x = \frac{- 2 \pm 4 i}{2}

.
⇒

x = \frac{2 (- 1 \pm 2 i)}{2}

∴ x = - 1 \pm 2 i

.
So, the two factors will be −1−2i, −1+2i
So, the product of linear factors of P(x) =

x^{4} + 4 x^{3} + 6 x^{2} + 4 x - 15

will
be (x−1)(x+3)(x+1+2i)(x+1−2i).

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[[1]] is polynomial P as a product of linear factors: P(x) = x4+4x3+6x2+4x-15[Hint 1 and−3 are zeroes of P]