Find without complete multiplication the coefficient of $x^{4}$ in the product of $5 x^{3} - 2 x^{2} + 3 x - 4$ and $- 4 x + 3 x^{2} - 2 x^{3}$ .

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Solution

The correct option is D $- 32$
$(5 x^{3} - 2 x^{2} + 3 x - 4) (2 - 4 x + 3 x^{2} - 2 x^{3}) = - 2 x^{3} \times 3 x + 8 x^{2} \times - 2 x^{2} + 5 x^{3} \times - 4 x$
$= - 6 x^{4} - 6 x^{4} - 20 x^{4} = - 32 x^{4}$

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Similar questions

Q. Match the following.

	Column - I		Column - II
(P)	$(x^{2} + 5) (x^{3} + 3) + 5$	(1)	${- x}^{3} - 3 x^{2} + 3 x + 2$
(Q)	$(\frac{- 10}{3} x y^{3}) \times (\frac{6}{5} x^{3} y)$	(2)	$- x^{3} + x^{2} + 3 x - 6$
(R)	$(x^{3} - x^{2} - x - 2) - (2 x^{3} + 2 x^{2} - 4 x - 4)$	(3)	$x^{5} + 5 x^{3} + 3 x^{2} + 20$
(S)	$(x^{3} - x^{2} - x - 2) + (2 x^{2} - 2 x^{3} + 4 x - 4)$	(4)	$- 4 x^{4} y^{4}$

Q. The multiple roots of

x^{4} - 2 x^{3} - 11 x^{2} + 12 x + 36 = 0

are

Q. Identify polynomials in the following:

(i) f(x) = 4x³ − x² − 3x + 7

(ii) g(x) = 2x³ − 3x²+

\sqrt{x}

− 1

(iii) p(x) =

\frac{2}{3} x^{2} - \frac{7}{4} x + 9

(iv) q(x) = 2x² − 3x +

\frac{4}{x}

+ 2

(v) h(x) =

x^{4} - x^{\frac{3}{2}} + x - 1

(vi) f(x) =

2 + \frac{3}{x} + 4 x

Find the sum of the following polynomials and write the degree of the sum so obtained

(i) 2x³ $-$ 7x² + 3x + 4 ; 2x³ $-$ 3x² + 4x + 1
(ii) 3x² + 5x $-$ x⁷ ; $-$ 3x² + 5x + 8
(iii) x⁴ + 5x³ + 7x ; 4x³ $-$ 3x² + 5
(iv) y² + 2y $-$ 5 ; y³ + 2y² + 3y + 4 ; y³ + 7y $-$ 2
(v) 5m² + 3m + 8 ; m³ $-$ 6m² + 4m ; m³ $-$ m² $-$ m + 5

Identify polynomials in the following :

(i) f(x) = $4 x^{3} - x^{2}$ - 3x + 7

(ii) g(x) = $2 x^{3} - 3 x^{2} + \sqrt{x} - 1$

(iii) p(x) = $\frac{2}{7} x^{2} - \frac{7}{4} x + 9$

(iv) q(x) = $2 x^{2} - 3 x + \frac{4}{x} + 2$

(v) h(x) = $x^{4} - x^{\frac{3}{2}} + x - 1$

(vi) f(x) = 2 + $\frac{3}{x} + 4 x$

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Find without complete multiplication the coefficient of x4 in the product of 5x3−2x2+3x−4 and −4x+3x2−2x3.

The correct option is D −32(5x3−2x2+3x−4)(2−4x+3x2−2x3)=−2x3×3x+8x2×−2x2+5x3×−4x=−6x4−6x4−20x4=−32x4

Find without complete multiplication the coefficient of $x^{4}$ in the product of $5 x^{3} - 2 x^{2} + 3 x - 4$ and $- 4 x + 3 x^{2} - 2 x^{3}$ .

The correct option is D $- 32$
$(5 x^{3} - 2 x^{2} + 3 x - 4) (2 - 4 x + 3 x^{2} - 2 x^{3}) = - 2 x^{3} \times 3 x + 8 x^{2} \times - 2 x^{2} + 5 x^{3} \times - 4 x$
$= - 6 x^{4} - 6 x^{4} - 20 x^{4} = - 32 x^{4}$